MHB Find Center of Groups of Order 8

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To find the center of groups of order 8, it is established that if G/Z(G) is cyclic, then G equals Z(G). This implies that the order of Z(G) can only be 1, 2, or 8. However, Z(G) cannot be 1, as this would contradict the nature of abelian groups where Z(G) equals G. The discussion emphasizes the use of the class equation, which states that the order of G equals the order of Z(G) plus the sum of indices of normalizers for elements not in Z(G). Each index must divide 8 and be even, leading to the equation 8 = 1 + 2k, which must have integer solutions.
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How to find the center of groups of order 8?
 
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Fessenden said:
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 $1$?
 
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?
 

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