Let $G$ be a group of order $1331$. Prove that $G$ has at least $11$ elements of order $11$.
$|G|=1331=11^3$
So by First Sylow's theorem, there exists a Sylow $11$-subgroup of G.
By Third Sylow's theorem, the number of such subgroups is $11k+1$ and $11k+1|1331$, thus, this is only possible for $k=0$. This means that the Sylow $11$-subgroup is unique, and therefore there exist at least $10$ elements of order $11$ in $G$.
So it appears I'm missing one element to complete the proof. Have I done something wrong?
Best Answer
I agree with the remark of David Wheeler - in general, if $G$ is a group with $|G|=p^n$, $p$ prime, then $|\{g \in G : g^p=1 \}| \geq p$. This follows basically from Cauchy's Theorem.
By the way, in general it holds that $|\{g \in G : order(g)=p \}| \equiv -1$ mod $p$. This can be read off from the famous proof of James McKay (see for example here) of Cauchy's Theorem.