Free products of finite groups acting on regular rooted trees
Let finite number of finite groups be given. Let n be the largest order of their composition factors. We prove explicitly that the group of finite state automorphisms of rooted n-tree contains subgroups isomorphic to the free product of given groups.
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Інститут прикладної математики і механіки НАН України
2007
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| Цитувати: | Free products of finite groups acting on regular rooted trees / C.K. Gupta, N.D. Gupta, A.S. Oliynyk // Algebra and Discrete Mathematics. — 2007. — Vol. 6, № 2. — С. 91–103. — Бібліогр.: 11 назв. — англ. |
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irk-123456789-1573542019-06-21T01:30:17Z Free products of finite groups acting on regular rooted trees Gupta, C.K. Gupta, N.D. Oliynyk, A.S. Let finite number of finite groups be given. Let n be the largest order of their composition factors. We prove explicitly that the group of finite state automorphisms of rooted n-tree contains subgroups isomorphic to the free product of given groups. 2007 Article Free products of finite groups acting on regular rooted trees / C.K. Gupta, N.D. Gupta, A.S. Oliynyk // Algebra and Discrete Mathematics. — 2007. — Vol. 6, № 2. — С. 91–103. — Бібліогр.: 11 назв. — англ. 1726-3255 http://dspace.nbuv.gov.ua/handle/123456789/157354 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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Let finite number of finite groups be given. Let n
be the largest order of their composition factors. We prove explicitly that the group of finite state automorphisms of rooted n-tree
contains subgroups isomorphic to the free product of given groups. |
| format |
Article |
| author |
Gupta, C.K. Gupta, N.D. Oliynyk, A.S. |
| spellingShingle |
Gupta, C.K. Gupta, N.D. Oliynyk, A.S. Free products of finite groups acting on regular rooted trees Algebra and Discrete Mathematics |
| author_facet |
Gupta, C.K. Gupta, N.D. Oliynyk, A.S. |
| author_sort |
Gupta, C.K. |
| title |
Free products of finite groups acting on regular rooted trees |
| title_short |
Free products of finite groups acting on regular rooted trees |
| title_full |
Free products of finite groups acting on regular rooted trees |
| title_fullStr |
Free products of finite groups acting on regular rooted trees |
| title_full_unstemmed |
Free products of finite groups acting on regular rooted trees |
| title_sort |
free products of finite groups acting on regular rooted trees |
| publisher |
Інститут прикладної математики і механіки НАН України |
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2007 |
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http://dspace.nbuv.gov.ua/handle/123456789/157354 |
| citation_txt |
Free products of finite groups acting on regular rooted trees / C.K. Gupta, N.D. Gupta, A.S. Oliynyk // Algebra and Discrete Mathematics. — 2007. — Vol. 6, № 2. — С. 91–103. — Бібліогр.: 11 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
| work_keys_str_mv |
AT guptack freeproductsoffinitegroupsactingonregularrootedtrees AT guptand freeproductsoffinitegroupsactingonregularrootedtrees AT oliynykas freeproductsoffinitegroupsactingonregularrootedtrees |
| first_indexed |
2025-07-14T09:47:53Z |
| last_indexed |
2025-07-14T09:47:53Z |
| _version_ |
1837615257004015616 |
| fulltext |
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.
Algebra and Discrete Mathematics RESEARCH ARTICLE
Number 2. (2007). pp. 91 – 103
c© Journal “Algebra and Discrete Mathematics”
Free products of finite groups acting on
regular rooted trees
C. K. Gupta, N. D. Gupta, A. S. Oliynyk
Dedicated to V.I. Sushchansky on the occasion of his 60th birthday
Abstract. Let finite number of finite groups be given. Let n
be the largest order of their composition factors. We prove explic-
itly that the group of finite state automorphisms of rooted n-tree
contains subgroups isomorphic to the free product of given groups.
1. Introduction
In last time many authors pay attention to automorphism groups or
rooted tree. There are two main reasons for it. First, these groups help
to obtain many deep purely algebraic results. Second, in studying auto-
morphism groups of rooted trees many interesting connections between
algebra, automata theory, dynamical systems, functional analysis etc.
arise. For detailed overview see [GNS] and references therein.
We continue in the presented paper to study the subgroup structure
of the automorphism group of a rooted tree. We consider free products
of finite groups acting on regular rooted trees by the so-called finite state
automorphisms. This is a natural question, since in the automorphism
group of a regular rooted tree “almost all” subgroups are free [3, 4] and
free products are rich in free subgroups. For previous results on free
products acting on rooted trees see [5, 6, 2, 7, 8].
The main result of our paper is the following sufficient condition of
existence of a faithful action on the regular rooted tree Tn by finite state
automorphisms.
The paper was mainly written during the visit of third author to the University of
Manitoba. The third author express his gratitude for this Institution for its hospitality.
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Theorem 5.2. Let H1, . . . , Hk be finite groups and suppose that the
orders of all composition factors of each Hi are bounded above by n. Then
the free product H1 ∗ · · · ∗Hk acts faithfully on the regular rooted tree Tn
by finite state automorphisms.
The work is organized as follows. In Section 2 all necessary definitions
and notations are given. In Section 3 we give some necessary conditions
of existence of a faithful action by finite state automorphisms on Tn of a
given free product of finite groups. The section 4 is the main part of the
paper. It contains an explicit construction used in the proof of the main
result and of the corollaries in Section 5. Section 6 contains an example
of a faithful action, constructed by the methods of Section 4. And the
last Section 7 is devoted to some open question concerning free products
and groups of finite state automorphisms of regular rooted trees.
2. Preliminaries
For more details on the contents of this section see, for example, [GNS].
Let X be a finite alphabet, |X| = n ≥ 2. Denote by X∗ the free monoid
generated by X. In other words, X∗ is the set of all finite words over the
alphabet X, including the empty word Λ, with the operation of concate-
nation. We identify X∗ with the disjoint union of the Cartesian powers
of X
⋃
i≥0
Xi,
where X0 = {Λ}. The length |v| of a word v ∈ X∗ is i ≥ 0 such that
v ∈ Xi. Also consider the set XN of all sequences (or the so called
ω-words) of the form
x1x2x3 . . . ,
where xi ∈ X, i ≥ 1.
Define a partial order “<” on X∗ by the rule
for u, v ∈ X∗ u < v iff u = vv1for some v1 ∈ X∗, v1 6= Λ.
Then the diagram of the poset (X∗, <) is a regular rooted tree denoted
Tn. The set of vertices of Tn is X∗, Λ is the root and two vertices u, v are
connected if and only if for some x ∈ X we have u = vx or v = ux. All
vertices of this tree are naturally partitioned into levels, where the ith
level is Xi, i ≥ 0. Each vertex from the ith level (i ≥ 0) is adjacent with
n vertices from the (i+ 1)st level and with one vertex from the (i− 1)st
level, except for i = 0.
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.C. K. Gupta, N. D. Gupta, A. S. Oliynyk 93
Figure 1: Regular rooted tree T3, constructed for the alphabet X =
{0, 1, 2}
Denote the automorphism group of the rooted tree Tn by GAn. Ev-
ery automorphism of Tn fixes the root Λ. This means, that for given
automorphism f ∈ GAn, word u ∈ X∗ and x ∈ X we have
(ux)f = ufxπ(u),
where π(u) is a permutation from the symmetric group Sn, depending
only on the word u. This property leads us to a definition of the action
of GAn on XN. Namely, for f ∈ GAn and w = x1x2x3 . . . xn . . . ∈ XN,
then
wf = xπΛ
1 x
πx1
2 x
πx1x2
3 . . . x
πx1x2...xn−1
n .
We obtain in this way the permutation group (GAn, X
N). As a permu-
tation group this group is isomorphic to the infinitely iterated wreath
product of symmetric groups Sn:
(GAn, X
N) ≃
(
∞
≀
i=1
Sn, X
N
)
.
In particular, this means that for arbitrary k ∈ N each automorphism
f ∈ GAn can be written as a pair
f = (fk, f
k),
where fk ∈ ≀
k
i=1
Sn and fk : Xk → ≀
∞
i=k+1
Sn, i.e. for every u ∈ Xk
we have fk(u) ∈ GAn. We call the automorphism fk(u) the state of the
automorphism f in u. In particular, f(Λ) = f so that f is a state of itself.
An automorphism f ∈ GAn is called finite state if it has only finitely
many different states. All finite state automorphisms form a subgroup of
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GAn that is called the finite state automorphism group and is denoted
by FGAn. This group contains a proper subgroup FinGAn of finitary
automorphisms. An automorphism f ∈ GAn is called finitary if there
exist k ≥ 0 such, that for all u ∈ Xk the state fk(u) is the identity.
In other words, f does not change the mth letter in every (finite or ω-)
word over X for all m > k. This group is locally finite since it can be
decomposed as the direct limit of wreath products ≀
k
i=1 Sn, k ≥ 1 with
the natural embeddings.
3. Some necessary conditions
Proposition 3.1. Let H1, . . . , Hk (k ≥ 2) be finite subgroups of the
finite state automorphism group FGAn. All composition factors of Hi
(1 ≤ i ≤ k) are subgroups of Sn.
Proof. We will prove that for some m ≥ 1 the group H1 is embeddable
into ≀
m
i=1
Sn. As a subgroup of FGAn the group H1 acts on the levels
of Tn. Suppose that all of these actions are non-faithful. Note that the
kernel of the action on the lth level contains the kernel of the action on
(l + 1)-st level, l ≥ 1. Since H1 is finite, the series of kernels have to
stabilize on some nontrivial subgroup of H1. This means that the action
of H1 on Tn is not faithful which contradicts the selection of H1. Hence
for somem ≥ 1 the groupH1 acts faithfully on the set Xm. This precisely
means embeddability of H1 into the wreath product ≀
m
i=1
Sn. As it follows
from [9], this implies the statement of the proposition.
Remark 3.2. Note, that this condition for finite group H is also sufficient
for being a subgroup of FGAn.
Proposition 3.3. Suppose that the group G splits into a free product:
G = H1 ∗ · · · ∗Hk
of finite groups. Then at most one of the groups H1, . . . , Hk is a subgroup
of FinGAn.
Proof. Assume that two groups (for example H1, H2) are contained in
FinGAn. It follows immediately from the local finiteness of FinGAn
that the group generated by them is finite. This is a contradiction.
4. Sufficient condition
We start with two useful lemmata about wreath products.
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.C. K. Gupta, N. D. Gupta, A. S. Oliynyk 95
Let (G,M) be a transitive permutation group and let ≀
m
i=1(G,M) be
its (permutational) wreath power. For m ≥ 2, the elements of ≀
m
i=1
(G,M)
will be written as pairs (g, f(x)), where g ∈ ≀
m−1
i=1
(G,M) and f(·) :
Mm−1 → G. We write h instead of f(x) in case f(x) ≡ h for some
h ∈ G. Let (H,N) be a subgroup of ≀
m
i=1
(G,M). The canonical embed-
ding
τ : (H,N) →֒
m+1
≀
i=1
(G,M)
is given by
τ(h) = (h, e), h ∈ H.
Here e denotes the identity element of H. Also define the j-diagonal (for
j ≥ 0) embedding
ψm,j : (H,N) →֒
j
≀
i=1
(G,M).
Here ψ0 is the identity mapping and for j > 0
ψj(h) = (ψj−1(h), h), h ∈ H.
Lemma 4.1. Let (H,N) be a subgroup of the wreath power ≀
m
i=1
(G,M).
Then for every j ≥ 1 the wreath power ≀
m+j
i=1
(G,M) contains a subgroup
(H1, N1) isomorphic to (H,N) as a permutation group.
Proof. Fix an arbitrary m ∈M . Define the subset
N1 = N × {m} × · · · × {m}
︸ ︷︷ ︸
j times
⊂Mm+j .
Denote by H1 the image of H under τ j , the jth iteration of the canonical
embedding. Then (H1, N1) is the necessary subgroup.
Lemma 4.2. Let (H,N) be a subgroup of the wreath power ≀
m
i=1
(G,M).
Then for every w ∈ Mm there exists a subgroup (H1, N1) of ≀
m
i=1
(G,M)
isomorphic to (H,N) as a permutation group, such that w ∈ N1.
Proof. Consider the case w 6∈ N . Chose an arbitrary u ∈ N . Since the
group (G,M) is transitive, so is the group ≀
m
i=1
(G,M). Hence it contains a
permutation π such that π(u) = w. Then (π−1Hπ, π(N)) is the required
group.
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Consider now finite groups H1, . . . , Hk such that all the composition
factors of Hi (1 ≤ i ≤ k) are subgroups of Sn. This means that Hi is a
subgroup of
(
≀
mi
i=1
Sn, X
mi
)
for some mi ∈ N and 1 ≤ i ≤ k. Suppose
additionally that Hi acts regularly on some subset Ai ⊆ Xmi , 1 ≤ i ≤
k. Using Lemma 4.1 we may assume that m1 = . . . = mk = m. By
Lemma 4.2 we have that the intersection
⋂m
i=1Ai is not empty. Denote
some common element by w.
Define now for every i, 1 ≤ i ≤ k, the embedding
ϕi : Hi →֒
mk
≀
i=1
Sn
as the composition ϕi = ψi · τ
m−i. This means that at first we take the
i-diagonal embedding and then use the canonical embedding m− i times.
Consider the following k subsets of the set Xmk:
M1 = {uww . . . w : u ∈ A1, u 6= w}
M2 = {uuw . . . w : u ∈ A2, u 6= w}
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Mk = {uuu . . . u : u ∈ Ak, u 6= w}.
Lemma 4.3. The sets M1, . . . ,Mk are nonempty and pairwise disjoint.
Proof. Follows from nontriviality of the given groups and regularity of
their actions on the sets A1, . . . , Ak.
It is easy to see that
Mi = {(w . . . w
︸ ︷︷ ︸
k times
)ϕi(h) : h ∈ Hi, h 6= e}, 1 ≤ i ≤ k.
Now define the following subsets of the set Xmk:
Di =
⋃
1≤j≤k
j 6=i
{vϕi(h) : v ∈Mj , h ∈ Hi}, 1 ≤ i ≤ k.
Lemma 4.4. The set Di, 1 ≤ i ≤ k is a union of the orbits of the action
of ϕi(Hi) on Xmk.
Proof. Let u ∈ Di. Then for some h1 ∈ Hi, h2 ∈ Hj , h2 6= e, j 6= i, we
have
u = ((wk)ϕi(h2))ϕi(h1).
This implies the required assertion.
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.C. K. Gupta, N. D. Gupta, A. S. Oliynyk 97
Note that neither Mi, nor Di (1 ≤ i ≤ k) contains the word wk.
We will use a presentation of GAn as a wreath product
GAn ≃
mk
≀
i=1
(Sn, X)
≀
(
∞
≀
i=1
(Sn, X)
)
≃
mk
≀
i=1
(Sn, X)
≀GAn.
Define finally for each i (1 ≤ i ≤ k) two maps fi1, fi2 : Hi −→ GAn.
Namely, for arbitrary element h ∈ Hi denote fi1(h) by h′ and fi2(h) by
h′′. Then as elements of the wreath product above they have the form
h′ = (e, h̄), h′′ = (ϕi(h), h̄)
and the map h̄ : Xmk −→ GAn acts by the rule
h̄(v) =
{
h′′, if v ∈ Di
h′, otherwise
.
Then for every v ∈ XN presented in the form v = v1v2v3 . . ., where
vi ∈ Xmk, i ≥ 1, we get:
vh′
= ve
1(v2v3 . . .)
h̄(v1),
vh′′
= v
ϕi(h)
1 (v2v3 . . .)
h̄(v1).
Proceeding this way we have
vh′
= ve
1v
π1
2 vπ2
3 . . . ,
vh′′
= v
ϕi(h)
1 vπ1
2 vπ2
3 . . . ,
where
πj =
{
ϕi(h), if vj ∈ Di
e, otherwise
, j ≥ 2.
Lemma 4.5. The maps fi1, fi2 are faithful representations of Hi in
FGAn.
Proof. Let h ∈ Hi. The sets of states of the automorphisms h′ and h′′
are equal to the set of states {h′(v), h′′(v) : |v| ≤ km}. This implies that
fi1, fi2 are maps into the group FGAn of finite state automorphisms.
Let h1, h2 ∈ Hi be arbitrary elements. For v = v1v2v3 . . . ∈ XN,
where vi ∈ Xmk, i ≥ 1, we have
(vh′
1)h′
2 = (ve
1(v2v3 . . .)
h̄1(v1))h′
2 = ve
1((v2v3 . . .)
h̄1(v1))h̄2(v1)
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and
(vh′′
1 )h′′
2 = (v
ϕi(h1)
1 (v2v3 . . .)
h̄1(v1))h′′
2 = v
ϕi(h1h2)
1 ((v2v3 . . .)
h̄1(v1))h̄2(v
ϕi(h1)
1 ).
By Lemma 4.4 h̄1(v1) and h̄2(v
ϕi(h1)
1 ) are equal to h′1 and h′2 or to h′′1 and
h′′2, respectively. Since ϕi is an embedding, we inductively obtain that
both fi1 and fi2 are faithful representations.
Denote by G1 and G2 the subgroups of FGAn generated by the images
f11(H1), . . . , fk1(Hk) and f12(H1), . . . , fk2(Hk), respectively.
Theorem 4.6. The groups G1 and G2 are isomorphic to the free product
H1 ∗ · · · ∗Hk.
We need for the proof the following generalization of the well-known
“ping-pong” lemma.
Lemma 4.7. [6] Let a permutation group G on a set Ω be generated by
its proper subgroups G1, . . . , Gm (m ≥ 2) and at least one of them has
order greater then 2. If there exist pairwise disjoint nonempty subsets
Ω1, . . . ,Ωm of Ω such that for i ∈ {1, . . . ,m} the next condition holds:
ωg ∈ Ωi for ω ∈ Ωj , j 6= i, and g ∈ Gi, g 6= 1,
then the group G splits into the free product
G = G1 ∗ · · · ∗Gm.
Proof of theorem 4.6. Let us prove our statement for the group G2.
The group G2 as a subgroup of FGAn is a permutation group act-
ing on the set XN. Define for the subgroups f12(H1), . . . , fk2(Hk) the
following subsets Ω1, . . . ,Ωk of XN:
Ωi = (Xmk)∗Miw
N =
= {u1 . . . ulvww . . . : l ≥ 0, u1, . . . , ul ∈ Xmk, v ∈Mi}, 1 ≤ i ≤ k.
Since w 6∈Mi, for 1 ≤ i ≤ k, the subsets are well defined.
By Lemma 4.3 the subsets Ω1, . . . ,Ωk are pairwise disjoint.
Consider arbitrary indices i, j (1 ≤ i, j ≤ k, i 6= j). Let u =
u1 . . . ulvww . . . ∈ Ωj , l ≥ 0, u1, . . . , ul ∈ Xmk, v ∈ Mj and h ∈ Hi.
We will show that ufi2(h) ∈ Ωi. Since Mj ⊂ Di and wk 6∈ Di using the
rule of action of fi2(h) on XN we can write the word ufi2(h) in the form:
u′1 . . . u
′
lv
′(wk)ϕi(h)ww . . . .
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.C. K. Gupta, N. D. Gupta, A. S. Oliynyk 99
Here u′1, . . . , u
′
l, v
′ are some elements of Xmk. This presentation implies
that ufi2(h) ∈ Ωi by the definition of the set Mi.
Applying Lemma 4.7 we immediately obtain a factorization
G2 = f12(H1) ∗ · · · ∗ fk2(Hk).
By Lemma 4.5 it means that
G2 ≃ H1 ∗ · · · ∗Hk
which completes the proof. �
5. Embedding theorems
In this section we present some interesting results following from the
construction of the previous section and in particular from theorem 4.6.
We start with the following corollary of the Kaloujnine-Krasner wreath
product embedding theorem [10].
Lemma 5.1. Let H be a finite group, let m be the length of its com-
position series and suppose that the orders of the composition factors of
H are less than or equal to n. Then for an alphabet X (|X| = n) the
wreath product
(
≀
m
i=1
Sn, X
m
)
contains a regular subgroup (Γ, A) with
Γ ≃ H.
Proof. Let
H = H1 ⊲H2 ⊲ · · · ⊲Hm ⊲ {1}
be some composition series of H. Due to the Kaloujnine–Krasner Theo-
rem H acts regularly on the set
(H1/H2) × (H2/H3) × · · · × (Hm−1/Hm) ×Hm
as a subgroup of the standard wreath product
(H1/H2) ≀(H2/H3) ≀ · · · ≀(Hm−1/Hm) ≀Hm.
Since by assumption all factors of the wreath product have size ≤ n it
naturally embeds into
(
≀
m
i=1
Sn, X
m
)
. This completes the proof.
Theorem 5.2. Let H1, . . . , Hk be finite groups and suppose that the
orders of all their composition factors are bounded above by n. Then the
free product H1 ∗ · · · ∗Hk acts faithfully on the regular rooted tree Tn by
finite state automorphisms.
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Proof. By Lemma 5.1 all given groups satisfy the conditions of the previ-
ous section. Use then the construction described there and Theorem 4.6
gives the required conclusion.
Corollary 5.3. Let n = 2, 3 or 4. Then finite groups H1, . . . , Hk are
embeddable into the group of finite state automorphisms FGAn if and
only if their free product H1 ∗ · · · ∗Hk is embeddable.
Proof. Sufficiency is obvious.
We have to prove necessity. By Proposition 3.1 all groups H1, . . . , Hk
have subnormal series with factors from the symmetric group Sn. But in
our case (S2, S3 or S4) it follows that the composition factors of all given
groups have size ≤ n. Now apply Theorem 5.2.
Corollary 5.4. Finite soluble groups H1, . . . , Hk are embeddable into
the group of finite state automorphisms FGAn if and only if their free
product H1 ∗ · · · ∗Hk is embeddable.
Proof. It is sufficient to prove necessity. By Proposition 3.1 all the groups
H1, . . . , Hk have subnormal series with factors from the symmetric group
Sn. This means that orders of these factors divide n!. Consider the
composition series which are refinements of the subnormal ones. Since
groupsH1, . . . , Hk are soluble their composition factors are cyclic of prime
order. It follows from above that their orders divide n!. Hence these
orders are bounded by n from above. The rest is to apply Theorem 5.2.
In case n = p is prime we have the natural Sylow p-subgroup of the
group GAp, namely the wreath product ≀
∞
i=1
Cp of infinity many copies of
the cyclic group of order p. Denote it by Syl(GAp). We can also consider
the finite state part of this group, i.e., the intersection FSyl(GAp) =
Syl(GAp) ∩ FGAp. It is easy to see that both groups Syl(GAp) and
FSyl(GAp) contain every finite p-group. As in the previous section,
for any finite p-groups H1, . . . , Hk we can use the completely analogous
construction and to prove the following
Theorem 5.5. Let H1, . . . , Hk be finite p-groups. Then the group
FSyl(GAp) contains subgroups isomorphic to the free product H1 ∗ · · · ∗
Hk.
6. An example
Let H1 = 〈a|a2 = 1〉 and H2 = 〈b|b3 = 1〉 be cyclic groups of orders 2
and 3, respectively. Then H1, H2 act regularly on the subsets A1 = {1, 2}
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.C. K. Gupta, N. D. Gupta, A. S. Oliynyk 101
Figure 2:
and A2 = {1, 2, 3} of X = {1, 2, 3} as the subgroups 〈(12)〉 and 〈(123)〉
of the symmetric group S3. Using the notation of Section 4 we have
n = 3, k = 2 and m = 1. Consider ϕ1(a) = ((12); e, e, e) and ϕ2(b) =
((123); (123), (123), (123)) (Figure 2).
Choose w = 1. Then we get M1 = {21}, M2 = {22, 33} and D1 =
{22, 33, 12}, D2 = {21, 32, 13}. Denote ai = f1i(a) and bi = f1i(b),
i = 1, 2. As elements of FGA3 these automorphisms have the following
recurrent form (Figure 3):
a1 = (e; a1, a2, a1, a1, a2, a1, a1, a1, a2)
a2 = (ϕ1(a); a1, a2, a1, a1, a2, a1, a1, a1, a2)
b1 = (e; b1, b1, b2, b2, b1, b1, b1, b2, b1)
b2 = (ϕ2(b); b1, b1, b2, b2, b1, b1, b1, b2, b1)
By Theorem 4.6 we have 〈ai, bi〉 = 〈ai〉 ∗ 〈bi〉 ≃ C2 ∗ C3, i = 1, 2.
Due to [11] the subgroups 〈b2i aib
2
i , aib
2
i aib
2
i ai〉 are free of rank 2.
7. Some open questions
Let us formulate some questions arising in connection with the obtained
results. Firstly, the question about possibility of omitting any additional
conditions in Theorem 5.2.
Question 7.1. Let finite groups H1, . . . , Hk be embeddable into FGAn.
Is their free product H1 ∗ · · · ∗Hk embeddable into FGAn?
In particular,
Question 7.2. Is the free product An ∗ An embeddable into FGAn for
n ≥ 5?
Here by An we denote the alternating group of degree n.
If the answer to Question 7.1 is negative then
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.102 Free products of finite groups...
Figure 3: Automorphisms a2 and b2
Question 7.3. For given finite groups H1, . . . , Hk compute the smallest
number n such that the free product H1 ∗ · · · ∗ Hk is embeddable into
FGAn.
By Theorem 5.2 this number does not exceed the order of the largest
composition factor of H1, . . . , Hk.
And finally a question about closeness under operation of free product
in the class of finite state automorphism groups.
Question 7.4. Let H1 ≤ FGAn and H2 ≤ FGAm (n,m ≥ 2). Is it true
that H1 ∗H2 < FGAk for some k ≥ 2?
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Contact information
C. K. Gupta Department of Mathematics and Astronomy
The University of Manitoba
R3T 2N2 Winnipeg
Canada
E-Mail: cgupta@cc.umanitoba.ca
N. D. Gupta Department of Mathematics and Astronomy
The University of Manitoba
R3T 2N2 Winnipeg
Canada
E-Mail: ngupta@ms.umanitoba.ca
A. S. Oliynyk Department of Mechanics and Mathematics
Kyiv Taras Shevchenko University
Volodymyrska 60
01033 Kyiv
Ukraine
E-Mail: olijnyk@univ.kiev.ua
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