Find Center of Groups of Order 8

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SUMMARY

The discussion focuses on finding the center of groups of order 8, emphasizing that if the quotient group \( G/Z(G) \) is cyclic, then \( G \) must equal \( Z(G) \). It establishes that the order of the center \( |Z(G)| \) can only be 1, 2, or 8, and highlights that \( |Z(G)| \) cannot be 1, as this would imply \( G \) is abelian and trivial. The class equation \( |G| = |Z(G)| + \sum_{a \not\in Z(G)} [G:N(a)] \) is introduced, noting that the indices must divide 8 and must be even, leading to the integer solution \( 8 = 1 + 2k \).

PREREQUISITES
  • Understanding of group theory concepts, particularly centers and quotient groups.
  • Familiarity with the class equation in group theory.
  • Knowledge of normalizers and centralizers in the context of group actions.
  • Basic algebraic manipulation and integer solutions in equations.
NEXT STEPS
  • Study the properties of cyclic groups and their centers.
  • Explore the implications of the class equation in various group orders.
  • Investigate the relationship between normalizers and conjugacy classes in groups.
  • Learn about abelian groups and their characteristics in relation to group centers.
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Mathematicians, particularly those specializing in abstract algebra, students studying group theory, and anyone interested in the structural properties of finite groups.

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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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