Let $G$ be a finite group with more than one element. Prove that $G$ has an element of prime order For this question why can't I just say since the identity is in every group, therefore the statement hold?
Here is what I have so far:
Let $g$ be an element of a group $G$, such that $g ≠ e$, and the order of $g$ is $n$. Let $p$ be some prime that divides $n$, then
$$g^n = (g^{n/p})^p = e.$$
Therefore the the statement is true.
Is this enough to prove the statement?
 A: Almost, you didn't prove that $g$ must have finite order, you also didn't prove that $g^{n/p}\neq e$. I think these are the only gaps, of course they are small gaps.
A: Let $G$ be a group of order $n$. Then every element of $G$ has order dividing $n$.
Let $g$ be a non-identity element with order dividing $n$, say $m$.
If $m$ is prime, then we are done.
If $m$ is composite, write $m=pq$, where $p$ is a prime.
Then $g^q$ is an element of order $p$ in $G$.
For your proof, you need to relate finiteness of $G$ to the order of $g$. Also, you need to show $|g^{n/p}|=p$ not just because $(g^{n/p})^p=e$.
A: Suppose $|a|=p_{1}^{n_{1}}p_{2}^{n_{2}}...p_{k}^{n_{k}}$.Then consider $a^{p_{1}^{n_{1}-1}p_{2}^{n_{2}}...p_{k}^{n_{k}}}$ has an order of $p_{1}$.
A: Let $a ∈ G$ be any element not equal to the identity. Suppose $|a| = n$. If
$n$ is prime we are done, otherwise by the Prime Factorization Theorem $n$ has a prime
divisor $p$. Suppose that $n = pk$. Then if $b = a^k$ we have that $b^p = a^{kp} = a^{n} = e$. So
$|b| ≤ p$. If $|b| = l < p$ then $e = b^l = a^{kl}$ which would contradict $|a| = n > kl$. Hence
the order of $b$ is $p$.
