# Thread: Suppose that |G| = p^n and further suppose that G is simple. Prove that |G| = p.

1. ## Suppose that |G| = p^n and further suppose that G is simple. Prove that |G| = p.

Suppose that $|G| = p^n$ and further suppose that $G$ is simple. Prove that $|G| = p$.

2. Originally Posted by joestevens
Suppose that $|G| = p^n$ and further suppose that $G$ is simple. Prove that $|G| = p$.

1) Using the Class Equation, show that any finite p-group has non-trivial center,

2) By induction on n, show that for any $0\leq k\leq n$ there exists a normal

sbgp. $N_k\leq G$

3) Solve your problem now.

Tonio

3. Originally Posted by joestevens
Suppose that $|G| = p^n$ and further suppose that $G$ is simple. Prove that $|G| = p$.
Or, use the fact that if a $p$-group has order $p^n$ then any subgroup of order $p^{n-1}$ must be normal.

4. Originally Posted by Drexel28
Or, use the fact that if a $p$-group has order $p^n$ then any subgroup of order $p^{n-1}$ must be normal.
You would also need to prove subgroups of order $p^{n-1}$ exist. This is not obvious (you need that all the composition factors in your composition series are of order p, or something like it).

5. Originally Posted by tonio
1) Using the Class Equation, show that any finite p-group has non-trivial center,
done.

Originally Posted by tonio
2) By induction on n, show that for any $0\leq k\leq n$ there exists a normal sbgp. $N_k\leq G$
done.

Originally Posted by tonio
3) Solve your problem now.

Tonio
n must be 1. then the only normal subgroups have order p and 1, which are $G$ and $\{e\}$.

6. Originally Posted by Swlabr
You would also need to prove subgroups of order $p^{n-1}$ exist. This is not obvious (you need that all the composition factors in your composition series are of order p, or something like it).
I'm sorry, isn't the fact that if $p^k\mid |G|$ where $|G|<\infty$ then $\exists H\leqslant G$ such that $|H|=p^k$ commonly known?

7. Originally Posted by Drexel28
I'm sorry, isn't the fact that if $p^k\mid |G|$ where $|G|<\infty$ then $\exists H\leqslant G$ such that $|H|=p^k$ commonly known?
Well, yes, but so is the fact that the centre is non-trivial. As the OP doesn't appear to know that the centre is non-trivial, we must assume they don't know that either! My point was, if one is going to prove one of maximal subgroups have order p', Z(G) is non-trivial', one would prove the latter.

Unless, of course, one doesn't know the class formula...

(I've used one' as if I'd used you' the last line is obviously false, because I'm pretty sure you know the class formula...so I was trying to not insult you, not trying to sound posh...)