3
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I know this is question has been asked several times on here, only hints given ,but just want to check if I have the right idea.

My attempt: Suppose G is non abelian finite group and $|Z(G)| \gt \frac {1}{4} |G|$. Since Z(G) is a subgroup of G then it’s order must divide that of G , i.e it could be of size |G|/3 , |G|/2 or |G| .That means the quotient group G/Z has order 1 ,2 or 3 . If equal to 1 that implies Z(G)=G so abelian , and also since 2 and 3 are prime numbers that means G/Z is cyclic which also implies it’s abelian. Which leads to a contradiction. Is that right? Thanks

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    $\begingroup$ Yeah, that's right. Indeed the minimum size of the inner automorphism group when it is non-trivial is $V$ - see here. $\endgroup$ – Parcly Taxel May 26 at 1:37
  • $\begingroup$ @Parcly Taxel thanks , also one quick question could one just define any quotient group as simply the left coset ? $\endgroup$ – Eden Hazard May 26 at 1:46
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    $\begingroup$ If the quotient is defined in the first place, then yes, since left coset equals right coset for normal subgroups. $\endgroup$ – Parcly Taxel May 26 at 1:47
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    $\begingroup$ Also, it is because if $G/Z(G)$ is cyclic, then $G$ is abelian. $\endgroup$ – Hongyi Huang May 26 at 3:55
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    $\begingroup$ The last part is not very clear. If $G/Z$ is cyclic, then of course it is abelian, but more importantly, it implies $G$ is abelian. $\endgroup$ – verret May 26 at 8:08

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