Commutator subgroup and center

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SUMMARY

The center of a group does not always contain the commutator subgroup, as demonstrated in the discussion. The theorem states that if N is a normal subgroup of G, then G/N is abelian if and only if the commutator subgroup C is contained in N. The center is indeed a normal subgroup, but there are instances, such as with the alternating group A_5, where the quotient G/Z(G) is not abelian, confirming that the center does not necessarily include the commutator subgroup.

PREREQUISITES
  • Understanding of group theory concepts, specifically commutator subgroups and centers.
  • Familiarity with normal subgroups and their properties.
  • Knowledge of quotient groups and their relationship to abelian groups.
  • Basic understanding of simple groups, particularly the alternating group A_5.
NEXT STEPS
  • Study the properties of commutator subgroups in various group structures.
  • Explore the implications of normal subgroups on the structure of quotient groups.
  • Investigate examples of simple groups and their centers, focusing on A_5.
  • Learn about the relationship between abelian groups and their quotient structures.
USEFUL FOR

Mathematicians, particularly those specializing in abstract algebra, students studying group theory, and anyone interested in the properties of commutator subgroups and centers in group structures.

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Homework Statement


Please confirm that the center of a group always contains the commutator subgroup. I am pretty sure its true.


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The Attempt at a Solution

 
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It's not. That would imply commutators commute, which there's no good reason to expect, and it should be easy to find a counterexample.
 
But my book has a theorem that says:

"If N is a normal subgroup of G, then G/N is abelian if and only if C is contained in N"
where C is the commutator subgroup.

Clearly the center is normal and its quotient group is abelian!
 
G/Z(G) is not necessarily abelian, e.g. Z(A_5) is trivial (since A_5 is simple), whence A_5/Z(A_5) =~ A_5, a non-abelian group.
 

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