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$G$ is the order $55$ group described by $G=\langle x,y \mid x^{11}=y^5=1, yxy^{-1}=x^4\rangle$. I am tasked with showing $|G/C(G)|=5$, but I don't know what $C(G)$ is supposed to be. It can't be the centre, as the centre is trivial - we need $|C(G)|=11$, but I cant think of what else it could be. Any advice?

Gentleman_Narwhal
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    I think the best thing would be to ask the person who gave that exercise. And if it's from a book, the notation should appear there. Maybe that's the derived (commutator) subgroup of $G$, but it's just my guess. – Mark May 12 '22 at 13:21
  • @Mark It's from a past paper from several years ago.... so I've no idea who wrote it. – Gentleman_Narwhal May 12 '22 at 13:26
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    Well, the derived subgroup of $G$ is indeed equal to $\langle x\rangle$, so it has order $11$. So maybe my guess was right. – Mark May 12 '22 at 13:29
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    The only elements of order $11$ are powers of $x$, so whatever $C(G)$ should mean, it must compute to $\left< x \right>$. – Adayah May 12 '22 at 13:30
  • Yeah I think derived/commutator subgroup makes sense, especially given the notation $C$. – Gentleman_Narwhal May 12 '22 at 13:32
  • $G$ is a group of order $pq$, with $p\mid q-1$. So it is either the cyclic group $C_{pq}$, or a non-abelian semidirect product, see here. So you can write down all subgroups you need, and your group is the nonabelian one. – Dietrich Burde May 12 '22 at 19:14

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