From: "A. Caranti"
Subject: Re: Group Theory question
Date: Thu, 23 Dec 1999 18:29:14 +0100
Newsgroups: sci.math
Keywords: finite groups with distinct but isomorphic characteristic subgroups
Fred Galvin wrote:
> Can you give an example of a finite group with two different
> characteristic subgroups that are isomorphic to each other?
In the literature there are several examples of finite p-groups (groups
of order a power of a prime p), constructed for various purposes, that
can be used to give examples of this kind.
To quote just one of the many possible examples, for each odd prime p
there are groups G of order p^10 with a center Z(G) of order p^6 such
that Z(G) is elementary abelian (that is, Z(G) is abelian, and every
element except the identity has order p), and the group Aut(G) of
automorphisms of G induces the identity on Z(G). Thus all subgroups of
Z(G) are characteristic. And any two subgroups of Z(G) of the same order
are clearly isomorphic. But of course Z(G) has plenty of distinct
subgroups of any order p, p^2, ..., p^5.
If anybody is interested, I can post a few references.
Andreas
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From: Fred Galvin
Subject: Re: Group Theory question
Date: Thu, 23 Dec 1999 14:33:15 -0600
Newsgroups: sci.math
On Thu, 23 Dec 1999, Fred Galvin wrote:
> Thanks! Someone else sent me an example by email: A_3+S_3
I mean G = A_3xS_3, i.e., the subgroup of S_6 generated by
{(1,2,3),(4,5),(4,6)}.
> has two characteristic subgroups of order 3.
Namely, one is the center of G, the other is contained in the subgroup of
G generated by the elements of order 2.
> Can a finite
Infinite groups are too easy, every subgroup of (Z,+) is characteristic.
> Abelian group have two isomorphic characteristic subgroups?
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From: Allan Adler
Subject: Re: Group Theory question
Date: 26 Dec 1999 04:35:13 -0500
Newsgroups: sci.math
I think I have made a little progress.
First, let G = Z/8Z + Z/2Z. Let H be the subgroup generated by (2,0)
and let K be the subgroup generated by (2,1). Then both H and K are
cyclic of order 4. I claim both are characteristic subgroups of G.
Proof: Let x=(1,0) and y=(0,1). Then 2x, -2x, 2x+y and -2x+y are
the only elements of order 4 in G. Let f be an automorphism of G.
Let f(x)=(a,b). Then f(2x)=(2a,0) = 2x or -2x, so f leaves H invariant.
Since f must permute the elements of order 4 of G, and since we have
just shown it permutes 2x and -2x among themselves, it must also
permute 2x+y and -2x+y among themselves, which proves that f must
also leave K invariant.
Therefore, the answer to Fred Galvin's question is, yes, an abelian
group can have two isomorphic but distinct characteristic subgroups.
The following analysis, if correct, will show that if G is an abelian
p-group with two distinct isomorphic characteristic subgroups then p
is equal to 2. There are also some other conditions. After that, I'll
add some miscellaneous remarks about counting the number of characteristic
subgroups of G.
I have improvised some terminology. In doing so, I hope I haven't pre-empted
existing terminology.
Let G be a finite abelian p-group. For nonnegative i,j, let G(i,j) denote
the subgroup of G consisting of elements of p^j G annihilated by p^i.
Lemma 0: Every automorphism of pG extends to an automorphism of G.
Proof: (There is probably a better way to say this, but I think this works.
Frattini subgroups are defined in Gorenstein's book Finite Groups, p.173,
but I'm not sure where the facts that I use about them are proved.)
Write G as a direct sum of cyclic groups. Let g1,g2,...,gr be
the generators of these cyclic groups. Let f be an automorphism of pG.
For i=1,...,r let hi = f(p gi). Let ki be an element of G such that
p ki = hi. Let F(gi) = ki for each i. Then F extends to a homomorphism
from G to itself. If none of the cyclic direct summands of G is cyclic
of order p, then F will be an automorphism of G, because F will induce
on G/pG the same endomorphism that f induces on pG/p^2G, these being the
Frattini quotients of G and pG respectively, and an endomorphism is an
automorphism iff it induces an automorphism on the Frattini quotient. On
the other hand, if some of the cyclic summands of G are of order p,
write G = G1 x G2, where G1 is the sum of the cyclic summands of order p
and G2 is the sum of the cyclic summands of order greater than p. On the
generators gi lying in G2, let F(gi)=ki as above, but choose ki in G2.
On G1, define F to be the identity automorphism. The Frattini argument
just given shows that F induces an automorphism of G2, so F gives an
automorphism of G.
Corollary: Every characteristic subgroup of G contained in pG is a
characteristic subgroup of pG.
The converse is obvious: every characteristic subgroup of pG is a
characteristic subgroup of G.
Lemma 1: Let H be a characteristic subgroup of G and assume that
H contains a cyclic direct summand of G of order p^e. Then H
contains G(e,0).
Proof: Let C be a cyclic direct summand of G contained in H and let x
be a generator of C. Write G as C x D. Let C have order p^e.
If y is an element of D such that p^e y = 0, then we define an
automorphism f of G by f(d)=d for d in D and f(x)=x+y. Since H contains
x, it contains all elements of D annihilated by p^e, and since H also
contains C, it follows that H contains all elements of G annihilated
by p^e, i.e. H contains G(e,0). QED.
Write G as a direct sum of cyclic groups. Let S be the set of
positive integers such that G has a cyclic direct summand of order p^i
and let a(i) be the number of such cyclic summands for i in S. Let K(i)
be the sum of the a(i) cyclic summands and let L(i) be the image of K(i)
in G/pG. Then G/pG is the direct sum of the Z/pZ-vector spaces L(i) with
i in S. By the projection of G into L(i), we mean the homomorphism from
G to L(i) given by the natural mapping of H to G/pG followed by the
projection of G/pG onto the direct summand L(i) of G/pG.
Lemma 2: Assume H is a characteristic subgroup of G not contained
in pG. Let e be the largest integer i such that the H does not lie in the
kernel of the projection of G onto L(i). Assume either that p>2 or a(e)>1.
Then H contains K(e) and, a fortiori, contains a cyclic direct summand of G
of order p^e.
Proof: Let x be an element of H whose image in L(e) is nonzero.
Then we can write x as u+v+w, where u belongs to K(e), v belongs to
the sum of all K(j) with j less than e and w belongs to the sum of all
pK(j) with j greater than e. Moreover, the image of u in L(e) will also
be nonzero. It is easy to prove that the group of automorphisms of K(e)
acts transitively on elements of L(e) which are nonzero modulo p L(e),
and since K(e) is a direct summand of G, such automorphisms extend to G.
Therefore, after applying such an automorphism we can assume that u is
actually one of the generators of one of the cyclic direct summands in
our decomposition of G. Applying the argument of Lemma 1, we can construct
an automorphism of G fixing K(j) for j not equal to e and sending u to u-v.
That automorphism therefore sends u+v+w to u+w. If p>2, apply an automorphism
of G sending u to itelf and w to -w, and then get u+w and u-w in H,
whence 2u is in H, so u is in H. If a(e)>1 and p=2 then we can find an
automorphism of G) permuting the cyclic summands of K(e) and sending u
to the generator t of another cyclic summand in the decomposition
of G and sending w to -w. Then u+w goes to t-w, so H contains the sum u+t of
u+w and t-w. Since u+t can be sent to u by an automorphism of K(e), the
conclusion of the Lemma still holds.
When H is a characteristic subgroup of G not contained in pG, we will
call the integer e the height of H in G. The integer e depends only
on G and H, not on the particular choice of cyclic subgroups in the
direct sum decomposition of G. We will call H a quasiregular characteristic
subgroup of G if either p>2 or a(e)>1, where e is the height of H.
Note that our definitions imply that a quasiregular characteristic
subgroup is not contained in pG. Moreover, if p=3, every characteristic
subgroup H of G is quasiregular, provided H does not lie in pG.
Lemma 3: Let H be a quasiregular characteristic subgroup of G of height e.
Then H contains G(e,0).
Proof: This follows at once from Lemmas 1 and 2.
By the depth of a characteristic subgroup H of G, we will mean the
largest integer d such that H lies in p^d G. The preceding lemmas
tell us about characteristic subgroups of depth 0. To study characteristic
subgroups of positive depth, we replace G by p^d G. We will say that
a characteristic subgroup H of depth d of G is semiregular if H is a
quasiregular characteristic subgroup of p^d G.
In the situation of Lemma 3, we can mod out G(e,0) from G and H by multiplying
by p^e. From p^e H in p^e G, we recover H as the preimage in G of p^e H under
multiplication by p^e. So we can try to proceed inductively.
One thing that can conceivably go wrong is that after modding out G(e,0),
the image of our quasiregular H can fail to be a semiregular characteristic
subgroup of p^e G. Even if we assume the image is semiregular, we will
get to mod out another subgroup of the form (p^e G)(i,j) and possibly
then wind up with a characteristic subgroup that is not semiregular,
and so on. We will say that a characteristic subgroup is regular if
it is semiregular and if these inductive procedures of modding out
always lead to semiregular characteristic subgroups.
When p>2, all characteristic subgroups of G are regular.
Lemma 5: Let G be an abelian p-group and let m,n,e be nonnegative integers.
Then (p^e G)(m,n) equals G(m,e+n) and the preimage of G(m,e+n) under
multiplication by p^e is G(m+e,n)+G(e,0).
Proof: An element x of G belongs to (p^e G)(m,n) iff (p^m x = 0 and x
lies in p^n (p^e G) = p^(e+n) G) iff x lies in G(m,e+n). As for the
second assertion, an element x of G lies in the preimage of G(m,e+n)
under multiplication by p^e iff ( p^m(p^e x)=0 and p^e x lies in
p^(e+n) G = p^e (p^n G) ) iff (p^(e+m) x = 0 and x differs from an
element of p^n G by an element in the kernel of multiplication by p^e.
QED.
Lemma 6: Assume p>2. Every characteristic subgroup H of G is a sum of
characteristic subgroups of G of the form G(i,j).
Proof: We prove this by induction on the order of G. It is obviously
true if G is trivial. If H has positive depth, we can replace G by pG
and again win by induction. Therefore, we can assume H has depth 0.
Then Lemma 3 implies that H contains G(e,0), where e is the height of G.
Multiplication by p^e maps H to the characteristic subgroup p^e H
of p^e G. Since e>0, p^e G has smaller order than G and therefore
our induction hypothesis applies. Therefore p^e H is the sum of
subgroups of p^e G of the form (p^e G)(m,n). By Lemma 5, we have
(p^e G)(m,n) = G(m,e+n) and the preimage of G(m,e+n) under
multiplication by p^e is G(m+e,n)+G(e,0). Therefore, H is a sum
of subgroups of the form G(i,j). QED.
Lemma 7: p>3. Let H,K be characteristic subgroups of G. If H and K
are isomorphic, then H=K.
Proof: If H and K both lie in pG, replace G by pG. Therefore, by induction
we can assume that one of H,K, say H, has depth 0. Let e be the height
of H. Then by Lemma 3, H contains G(e,0), i.e. H contains all elements of
G annihilated by p^e. Since H and K are isomorphic, H(e,0) has the same
order as K(e,0), so actually K contains G(e,0) also. Mod out G(e,0)
and win by induction on the order of G. QED.
I promised some miscellaneous remarks about counting the characteristic
subgroups of G. The following remarks apply to p>3 and to counting
regular characteristic subgroups when p=2. I'll have to do this quickly,
more or less off the top of my head, and can't be very careful about it.
Refer to the beginning of this article for the definition of the set S.
To S, we can associate a Young frame F. If S consists of the decreasing
sequence s(1) > s(2) > s(3) > ... > s(k), then the boxes in S are described
by ordered pairs (i,j) with 1 <= i <= k and 1 <= j <= s(i). I picture a
Young frame as sitting in the 1st quadrant with its sides flush against
the postive x-axis and positive y-axis, the corrdinate associated to a
given unit square box being the coordinates of its upper right corner.
By the 1st layer of the Young frame F, I mean the set of boxes of F associated
to coordinates (i,s(i)). By the r-th layer, I mean the set of boxes of
F associated to coordinates of the form (i,s(i)-r).
When H is a characteristic subgroup of G of depth 0 and height e, we can
associate to H the box of F associated to the coordinate (j,e), where
e=s(j). Thus, H gives us a box in the 0-th layer. More generally,
if H has depth d and has height e in p^d G, then we can associate to
H the box (j,e)=(j,s(j)-d) in the d-th layer. In the particular case
of the characteristic subgroup G(m,n) the box we get is associated
to the pair (i,j) where s(i)=m+n and j=m. When we take a gereral H,
written as a sum of subgroups G(m,n), we first make the sum irredundant
by eliminating summands contained in other summands. Then we get a set
of boxes in F. The set of boxes will form a sequence (i1,j1), (i2,j2),...
with strictly increasing levels and strictly increasing second coordinates.
Counting such sequences in a Young frame must be one of the uncountably
many combinatorial problems people have already posed about Young frames.
If someone knows how to count them, I'd be interested in hearing about it.
Allan Adler
ara@zurich.ai.mit.edu
****************************************************************************
* *
* Disclaimer: I am a guest and *not* a member of the MIT Artificial *
* Intelligence Lab. My actions and comments do not reflect *
* in any way on MIT. Morever, I am nowhere near the Boston *
* metropolitan area. *
* *
****************************************************************************
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From: wcw@math.psu.edu (William C Waterhouse)
Subject: Re: Group Theory question
Date: 28 Dec 1999 22:52:37 GMT
Newsgroups: sci.math
In article ,
Allan Adler writes:
> ...
> The following analysis, if correct, will show that if G is an abelian
> p-group with two distinct isomorphic characteristic subgroups then p
> is equal to 2.
> ...
> [most of the solution omitted]
> ...
Here is one idea which can simplify the treatment:
Lemma. Let G be a finite abelian p-group with p>2. Let H be
a characteristic subgroup. Let G = A x B, a direct product. Then H
is the direct product of its intersections with A and with B.
Proof. The projection of G onto A (say) is the sum of the two
automorphisms taking (a,b) to (2a, b) and to (-a, -b). As both
of those send H to itself, the projection takes H into H. Thus
the projections of H are the same as its intersections with A and B,
and that proves the result.
This of course allows us to write G as a direct product of cyclic
groups with H a direct product of subgroups in them.
Now say we have two of the cyclic factors, of order p^n and
of order p^k, with k less than or equal to n. Suppose we have an
element h of order p^s in H and . There is an automorphism
taking all generators by y to themselves and taking y to y+p^{n-k}x.
The image of h under that is still in H, as is its projection onto
; and that projection in will still have order p^s.
William C. Waterhouse
Penn State