1. ## p-groups

I have this Lemma: If $G$ is a finite abelian group whose order is divisible by a prime $p$, then $G$ contains an element of order $p$.

I have shown that for $x \in G$, $x \ne \{{1}\}$ and order of $x$ is $pm$, then $x^{m}$ is an element in $G$ of order $p$.
Now, suppose that $x$ has order $t$, where $(p,t)=1$. Since $G$ is abelian, $$ is a normal subgroup of G and $G/$ is an abelian group of order $|G|/t$. It is easy to see that $|G|/t$ is divisible by $p$ (since $|G|$ is divisible by $p$ and $(p,t)=1$) and is less than $|G|$.
Afterwards I know I need to show that there is an element in $G/$ whose order is $p$, but I'm stuck here. Please help me in completing the proof.

I have another corollary: A finite group $G$ is a $p$-group $\Longleftrightarrow$ $|G|$ is a power of $p$.

I have shown $( \Longrightarrow )$.
For $( \Longleftarrow )$, I used Lagrange's Theorem and shown that every subgroup of $G$ has order of power of $p$. How to continue the proof from here?

2. The corollary will based on the Lemma above and Cauchy's Theorem: If $G$ is a finite group whose order is divisible by a prime $p$, then $G$ contains an element of order $p$.

3. Originally Posted by deniselim17
I have this Lemma: If $G$ is a finite abelian group whose order is divisible by a prime $p$, then $G$ contains an element of order $p$.
do it by induction on the order of the group: if the order is 1, there's nothing to prove. so suppose |G| = n with n> 1, and the claim is true for any abelian group of order less than n.

let $p$ be a prime divisor of |G|. and choose $1 \neq x \in G.$ let $o(x)=m$ and $H = .$ if $\gcd(m,p)=1,$ then $p$ will divide |G/H| and obviously |G/H| < |G|. so by induction, G/H has

an element of order p, say $Hy$. so $y^p \in H.$ hence $y^{pm}=1,$ which gives you $o(y^m)=p,$ because $y^m \neq 1.$ why? if $p \mid m,$ then m = kp and $o(x^k)=p$ and you're done!

I have another corollary: A finite group $G$ is a $p$-group $\Longleftrightarrow$ $|G|$ is a power of $p$.

For $( \Longleftarrow )$, I used Lagrange's Theorem and shown that every subgroup of $G$ has order of power of $p$. How to continue the proof from here?
there's nothing left to continue! here's the definition of a p-group: a (finite or infinite) group is called a p-group if the order of every element of the group is a power of p.

Proof of the corollary: do induction again: if G is abelian, we're done. if there exists $x \in G - Z(G)$ such that p does not divide $[G:C_G(x)],$ then since p divides |G|, we must have $p \mid |C_G(x)|.$

since $|C_G(x)| < |G|,$ we're done by induction. if $p \mid [G:C_G(x)],$ for any $x \in G - Z(G),$ then $p \mid |Z(G)|,$ by the class equation, and we're done again because Z(G) is abelian. Q.E.D.

for another proof of Cauchy's theorem see here.

4. In following the proof by NonCommAlg by the conjugacy class equation the only step that needs work to be proven is that if $G$ is a finite abelian group and $p$ divides $|G|$ then there is an element of order $p$. Here is another way, though I perfer the one by NonCommAlg, involving the decomposition theorem. You can write $G \simeq C_1 \times ... \times C_n$ where $C_i$ are cyclic groups of prime order. Since $p$ divides $|G|$ it means $p$ divides some $C_i$. Therefore, $C_i \simeq \mathbb{Z}_{p^r}$. Let $H = p^{r-1}\mathbb{Z}_{p^r}$. Then $|H| = p$ and $H$ is cyclic therefore it is generated by an element of order $|H| = p$. This shows there is an element of order $p$.