Associated Topics || Dr. Math Home || Search Dr. Math

### 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
classes.

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
property:

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
http://mathforum.org/dr.math/
```
Associated Topics:
College Modern Algebra

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/