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What can we say about $|Z(G)|$ if $G$ is of order $3773 = 7^3 * 11$. Here $|Z(G)|$ means the size of the center.

This is an exercise in book about Sylow theorem and I have no idea what Sylow theorem has to do with size of center.

My idea is that center should be a normal subgroup, and according to Sylow theorem we know Sylow-11 and Sylow-7 subgroups are both normal subgroup. But I can't prove neither there are only two normal subgroup mentioned above, nor center should be which normal subgroup.

Shaun
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27rabbit
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2 Answers2

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Let $P$ be a $7$-Sylow subgroup, $Q$ is an $11$-Sylow subgroup. Then $P,Q$ are normal in $G$, $P\cap Q=\{e\}$ and by comparing cardinalities, $G=PQ$. It is well known that in this case we have:

$G\cong P\times Q\cong P\times\mathbb{Z_{11}}$

The center of a direct product is clearly the direct product of centers. The center of $\mathbb{Z_{11}}$ is obviously the whole group. Do you know what are the possible sizes of the center of a group of order $p^3$ where $p$ is a prime?

Mark
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  • Thanks!! I searched for group of order $p^3$ and got this https://math.stackexchange.com/questions/2272948/a-group-of-order-p3-and-its-quotient-with-the-center-of-the-group and http://mathonline.wikidot.com/the-center-z-g-of-a-nonabelian-group-of-order-p-3-is-p, now I understand the size of center of group of order $p^3$ should be either $p^3$ if group is abelian; or $p$ if not abelian. So here the size should be either 7 * 11 or 7^3 * 11, right? – 27rabbit Apr 08 '24 at 11:26
  • @27rabbit Yes, but you also need to verify that groups with centers of such sizes really exist, so you didn't list more options than there actually are. To get a center of size $7\times 11$ you just need to find a nonabelian group of order $7^3$. A classical example is the Heisenberg group over the field $\mathbb{F_7}$: https://en.wikipedia.org/wiki/Heisenberg_group – Mark Apr 08 '24 at 11:39
  • Thanks again! Got it, your explaination is very informative. – 27rabbit Apr 08 '24 at 11:51
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If $G$ is abelian, then you are done. So, let's suppose $G$ nonabelian. From $n_{11}=1$ follows that the Sylow $11$-subgroup is normal and, as such, it is union of conjugacy classes. Now, the conjugacy classes' size must divide the order of the group (orbit-stabilizer), and $11$ can't be "filled up" other than with $11=1+\dots+1$, as the other only option would be $11=7+1+1+1+1$, contradiction because $|P_{11}\cap Z(G)|$ can't have order $4$ (as $4\nmid |G|)$. Therefore, $P_{11}$ is central. But $|Z(G)|=11$ is ruled out because then the class equation would read: $7^3.11=11+7k+7^2l+7^3m$, for nonnegative integers $k,l,m$: contradiction, because $7\nmid11$. On the other hand, $|Z(G)|=7^2.11$ is ruled out because then there wouldn't be room between $Z(G)$ and the whole $G$ to accomodate the centralizers of the noncentral elements. Therefore, necessarily $|Z(G)|=7.11$ and $Z(G)\cong C_{77}$.

Kan't
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