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Subgroup and Order of a Group

Date: 11/06/2003 at 11:41:57
From: Lori
Subject: Modern Algebra (Subgroup And Order of a Group)

If o(G) = p^n, p is a prime number, and H is not equal to G and it is 
a subgroup of G, show that there exists an x that is an element of G 
and x is not an element of H such that x^(-1)Hx = H.

I find it confusing to figure out x^(-1)Hx = H for an element x.  If H 
is normal in G, then clearly x^(-1)Hx = H  for all x that are elements 
of G.  But then if H is just a subgroup of G, then x^(-1)Hx won't 
necessarily equal H unless proven.  I figure that o(G) = p^n is the 
reason for this to be true, but don't know if Sylow's theorem or 
something else needs to be used.

Thanks, Dr. Math has helped me figure out how to start a couple of
problems before and was very helpful.

Date: 11/07/2003 at 03:40:39
From: Doctor Jacques
Subject: Re: Modern Algebra (Subgroup And Order of a Group)

Hi Lori,

We need two results to prove this:

1) In a group of order p^n, where p is prime, the center of the group
   is not trivial.  This is proven by counting elements in conjugacy

2) The third isomorphism theorem: if f : G -> L is a homomorphism of
   groups, f defines a bijection between the subgroups of f(G) and
   the subgroups of G containing ker f.  A consequence of this is that
   if H1 and H2 are two subgroups of G containing ker f, and
   f(H1) = f(H2), then H1 = H2.

If you are not familiar with either of these facts, please write back.

Let G be a group of order p^n, p prime, and H be a proper subgroup of 
G.  Let C be the center of G.

If C is not a subgroup of H, this means that C contains an element x, 
not in H, that commutes with all elements of G, and, in particular, 
with all elements of H.  In that case, we have x^(-1)*h*x = h for all 
h in H, and we are done.

We may therefore assume that C is a subgroup of H.  We proceed by 
induction on n.

If n = 1, |G| = p, and the only proper subgroup of G is {e}--the 
result is trivial in this case.

Let us assume, by induction, that the theorem is true for all groups 
of order p^k, for k < n.

As C is normal in G, we may consider the canonical homomorphism
f : G -> G/C = L.

As C is not trivial, L is of order p^k with k < n, and the theorem 
holds in L by the induction hypothesis.  This means that there is an 
element b in L, not in f(H) such that:

  b^(-1)*f(H)*b = f(H)

(Note that f(H) is a proper subgroup of L, because otherwise we would 
have f(H) = f(G) and H = G by property (2) above.)

Now, b = f(a) for some a in G, and we have, by the homomorphism 

  f(a^(-1)*H*a) = f(H)

We can remark:

* a is not in H, because f(a) = b is not in f(H)

* C is a subset of a^(-1)*H*a, because, for any c in C, we have
  c = a^(-1)*c*a, and the latter is in a^(-1)*H*a, as C is a subset
  of H by hypothesis.

Can you conclude the proof?  Write back if you'd like to talk about 
this some more, or if you have any other questions.

- Doctor Jacques, The Math Forum 
Associated Topics:
College Modern Algebra

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