Re: (Group Theory) Is g^[G:A] necessarily in A if A isn't necessarily normal?

Snis Pilbor wrote: 
Hello,

     Fair warning:  this is not itself a homework problem but it IS
something which arose while working on a homework problem.

     Namely, if G a (maybe infinite) group, A a (maybe nonnormal)
subgroup of G w/finite index [G:A], is it true that for all g in G,
g^[G:A] is in A?

No. Let G=S_5, and let A be the point stabilizer of 5 (so you may think of A as S_4). In this case [G:A]=5. Then you
easily see that the permutation g=(12)(345) doesn't have the property g^5 \in A.

Anyway (as expected by your observation below), we do have
g^3 \in A.


     If A is normal, this is true and very trivial even.  But if A
isn't necessarily normal......  well, to ME it's hard, at least :)

     Some things I've tried...  g^i is in A for *some* 0<i<=[G:A].
Proof:  Assume not. Look at gA, g^2 A, ..., g^[G:A] A.  These are [G:A]
cosets, each can have one of [G:A]-1 values (since the coset 1A is by
assumption not in the list), by pigeonhole g^m A = g^n A with
0<m<n<=[G:A], so g^(n-m) in A contrary to assumption.


Looks good to me.


     We can define a multiplication on the elements g^i A, i integer,
by (g^i A)(g^j A)=g^(i+j) A, and it is seen to be well-defined and to
give rise to group structure (say A') even though A may not be normal.
We can define a map pi: <g> -> A' by pi(g^i)=g^i A and by isomorphism
theorems we get <g>/(Ker pi) ~ A', and of course Ker pi is <g>
intersect A so this could also be written <g>/(<g> intersect A) ~ A'...
but this rambling doesnt seem to lead anywhere..


True. C=<g> is cyclic, hence abelian, and hence all ITS subgroups (C intersect A in particular) are normal, so?

There are some well-studied cases, where the elements of another subgroup H form a set of representatives of left cosets of A (when this happens obviously the orders, if finite, match perfectly: |G|=|A| |H|). Since I'm no Derek Holt, I dare not give any general theory here :)

Cheers,

Jyrki Lahtonen, Turku, Finland
.

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