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- Thread starter Fessenden
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HINT: Start by proving that if $G/Z(G)$ is cyclic then $G=Z(G)$.How to find the center of groups of order 8?

This means that if $|G|=8$ then $|Z(G)|=1$, $2$ or $8$. Why can't it be $1$?

- Feb 15, 2012

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$|G| = |Z(G)| + \sum_{a \not\in Z(G)} [G:N(a)]$ where N(a) is the normalizer (or, in some texts, the centralizer) of the element a.

note each index [G:N(a)] must divide 8, and since we have the elements of Z(G) in the first term, and the a's are NOT in the center, each [G:N(a)] has to be either 2, or 4 (the only conjugacy classes of size 1 are elements of the center). in any case, each [G:N(a)] is an even number. is there an integer solution to:

8 = 1 + 2k?