Abstract algebra (If G is a group of order p^n p a prime number)

If G is a group of order p^n, p a prime number, then G contains
a normal subgroup of order p^(n-1).

Proof by induction on n.

This assertion is clearly true for n = 1.

Now, suppose this is true for all k < n.
Since G is a p-group, its center Z(G) is non-trivial. Thus, p | |Z(G)| by Lagrange's Theorem. By Cauchy's Theorem, Z(G) has an element of order p; call one such element g. Then, N = is normal in G, because any subgroup of Z(G) is normal in G. Now, consider G/N, which is of order p^(n-1). By inductive hypothesis, G/N has a normal H' of order p^(n-2). By the Correspondence Theorem, there exists a subgroup H of G containing N such that H/N = H'. Then, |H| = p^(n-1) and H is normal in G (again by Correspondence Theorem). So, G has a normal subgroup of order p^(n-1), and we are done.

I hope this helps!

my previous answer to this same question is here:

http://answers.yahoo.com/question/index;…

in reply, you referenced this question:

http://answers.yahoo.com/question/index;…

you seem to be concerned about the accuracy of my answer. although there are similarities the difference is this:

the second problem says this:

if K is a normal subgroup of G, and N is a normal subgroup of G containing K,

show N/K is normal in G/K.

my previous answer says this:

if L is a normal subgroup of G/H, (where H is a normal subgroup of H)

then L = K/H for some normal subgroup K of G.

now, there is no inconsistency between these two. kb's answer starts with

the group G, and then proves something about a factor group.

my answer starts with a factor group of G, and then proves something about G.

in fact the two answers are related:

for any homomophism G-->G/H there is a bijection between

subgroups of G/H and subgroups of G containing H given by:

N <---> N/H

moreover, N is normal in G if and only if N/H is normal in G/H.

i believe you have mis-interpreted my answer.

when i claim that if L is a normal subgroup of G/H,

then L = K/H for some normal subgroup of K, i first

prove that K = {g: gH is in L} is a subgroup. this needs to be shown

BEFORE one can even talk about K being a "normal subgroup".

it's circular reasoning to use the fact that K is normal to prove that

K is normal.

*****

the bijection N <-->N/H i mentioned earlier, is a straight-forward

consequence of the first isomorphism theorem. kb's post to another question of yours,

and my answer to this same question as you are asking now, are two sides

of this coin, they are converses of each other.