# Number of elements of order $11$ in group of order $1331$

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?

• The group $\mathbb{Z}/11\mathbb{Z}\times\mathbb{Z}/11\mathbb{Z}\times\mathbb{Z}/11\mathbb{Z}$ has more than one $11$-Sylow subgroups – egreg Jul 8 '15 at 15:56
• The third theorem tells you that there exists a unique $11^3$-Sylow subgroup, which is completely obvious. – egreg Jul 8 '15 at 16:01
• The question is wrong. The cyclic group of order $1331$ has only $10$ elements of order $11$. – Derek Holt Jul 8 '15 at 16:07
• By the way, it makes no sense to apply Sylow's Theorem to a group of order $p^n$ with $p$ prime, because it tells you nothing. – Derek Holt Jul 8 '15 at 16:09
• Are you sure the question does not say: prove $G$ has at least $11$ elements $g$ with $g^{11} = e$? Just sayin'. – David Wheeler Jul 8 '15 at 23:25

## 1 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.

• Cauchy is not even needed as the only possible element orders are powers of $p$, so one gets one of order $p$ by considering powers (and then the result is a basic fact about cyclic groups). – Tobias Kildetoft Jul 9 '15 at 8:19
• @Tobias, yes thanks. True but my second fact is of general nature. – Nicky Hekster Jul 9 '15 at 10:01