3
$\begingroup$

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!

$\endgroup$
0

1 Answer 1

3
$\begingroup$

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 $C_n$.

$\endgroup$
3
  • $\begingroup$ I'm sorry to ask this late, but: what's the $C_n$ group? $\endgroup$ Nov 26, 2014 at 17:24
  • $\begingroup$ 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. $\endgroup$ Feb 18 at 11:03
  • 1
    $\begingroup$ Funny, it's been over 8 years since I took my first algebra course. $\endgroup$ Feb 18 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.