# Show that for any group $G$ the index of the center in $G$ is not 51; $|G:Z(G)| \neq 51$.

I would like a hint on showing that $|G:Z(G)| \neq 51$, i.e. that the index of the center of a group is not 51. I have tried a couple different approaches;

1. We know that $G / Z(G) \cong \text{Inn}G$, and then by the Sylow theorems there exist sub-groups of $G$ of sizes 3 and 17 (exactly one each) since $51 = 3\cdot 17$. I have been unable to find a contradiction this way.

2. I also tried breaking it into cases on whether or not 17 divides $|Z(G)|$. When I assume $17 \not \mid |Z(G)|$ then I can derive a contradiction. But I cannot do the same for the case when $17 \mid |Z(G)|$.

3. I also tried breaking up the index: $|G:Z(G)| = |G:N_G(P)| \cdot |N_G(P) : C_G(P)| \cdot |C_G(P):Z(G)| = n_p \cdot |N_G(P)|$ where $N_G(P)$ is the normalizer and $C_G$ is the centralizer and $n_p$ is the number of Sylow $p$-groups (where $p$ is either 3 or 17), again this brought me nowhere.

I would appreciate any hint in the right direction.

Thanks!

And another hint: if $n$ is a natural number with gcd$(\varphi(n),n)=1$, then there is only one group of order $n$, which necessarily must be isomorphic to $C_n$.
• I'm sorry to ask this late, but: what's the $C_n$ group? Nov 26, 2014 at 17:24
• The cyclic group of order $n$ with a multiplicative group operation, same as / isomorphic to $\mathbb{Z}/n\mathbb{Z}$, with an additive group operation. Feb 18 at 11:03