At Least p+1 p-Subgroups in Finite Group G

[johng1]

Let p be a prime and G a finite group. Suppose G has more than one subgroup of order p. Then G has at least p+1 subgroups of order p. Notice the bound is sharp as shown by

\(\displaystyle G=\mathbb{Z}_p\oplus\mathbb{Z}_p\)

 

[caffeinemachine]

Let p be a prime and G a finite group. Suppose G has more than one subgroup of order p. Then G has at least p+1 subgroups of order p. Notice the bound is sharp as shown by

\(\displaystyle G=\mathbb{Z}_p\oplus\mathbb{Z}_p\)

Okay. But what is your question? :P

EDIT:

I can do it if we additionally assume that $G$ is abelian.
Let $H_1,\ldots,H_k$ be all the subgroups of order $p$.
Let $S=\{x\in G:x^p=e\}$.
Then it can be seen that $S=\bigcup_{i=1}^k H_i$.
Also, $S$ is a subgroup of $G$ since $G$ is abelian.

Also, no member in $S$ has order other that $1$ or $p$. Thus, by Cauchy's theorem, $|S|=p^m$ for some integer $m$.

Also notice that $H_i$ and $H_j$ intersect only in $\{e\}$ whenever $i\neq j$.

This leads to $p^m=(p-1)k+1$, giving $k=1+p+\cdots+p^{m-1}$.

So $k=1, 1+p,1+p+p^2,\ldots$, depending on the value of $m$.

Since it is given that $k>1$, we have $k\geq p+1$.

 

[johng1]

Oops, my bad. My proof was a variation of Wielandt's proof of the Sylow theorem that says the number of Sylow p subgroups is congruent to 1 mod p. I had forgotten that this proof actually shows the number of p subgroups of a given order is congruent to 1 mod p. (It is assumed that p divides the order of G.) So the problem is trivial from this result. My apologies.

(I had intended to give a hint by suggesting a search for Wielandt's proof. This search results in the full statement of the theorem.)