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Center of groups


New member
Apr 25, 2012
How to find the center of groups of order 8?


New member
Feb 21, 2012
How to find the center of groups of order 8?
HINT: Start by proving that if $G/Z(G)$ is cyclic then $G=Z(G)$.

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


Well-known member
MHB Math Scholar
Feb 15, 2012
another way to put this is: if G is abelian, then Z(G) = G (which isn't very interesting). i believe the hint Swlabr is trying to steer you to is the class equation:

$|G| = |Z(G)| + \sum_{a \not\in Z(G)} [G:N(a)]$ 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?