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By definition, the center of G is the set:

$Z(G)$ = {$g\in G|g^{-1}xg=x$ for all $x\in G$}

We need to show that:

  • The identity element exists
  • It is closed under the operation
  • For every element $g$, there exists in $G$ the inverse element $g^-1$

My attempt:

  • Since it is defined that:

$g^{-1}xg=x$

Thus implying that $g^{-1}\in G$.

  • For the identity element $e$ and any element $x\in G$, since the identity element is its own inverse, it then implies:

$e^{-1}xe=x$

Hence $e\in G$.

  • Now by multiplying by $g$ on both sides, we get:

$g(g^{-1}xg)=gx$

$(gg^{-1})xg=gx$

$(e)xg=gx$

$xg=gx$

Thus, it is closed under the operation, hence implying that $Z(G)$ is a subgroup of $G$. Your help and feedback would be really appreciated.

Ittay Weiss
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Sai82
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    http://www.proofwiki.org/wiki/Center_of_Group_is_Subgroup – grantfgates May 02 '14 at 07:52
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    I suggest you begin by writing down exactly what is meant by the statement "$Z(G)$ is closed under the group operation". Can I emphasize: exactly: not "well it kinda means something like. . .". If you don't know, look it up in your textbook or lecture notes. – David May 02 '14 at 07:59
  • It is not essential, but I would recommend definition $Z(G)={g\in G\mid xg=gx \text{ for all }x\in G}$. It is a good custom to avoid (writing of) inverses if that is possible. – drhab May 02 '14 at 09:20

1 Answers1

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Your proof requires more work. Try to understand first, for instance, what do you need to show about $e$ in order to conclude that it is in the center. You also need to show that the center is closed under the operation and under taking inverses. Try to be very explicit about what you need to show in order to establish that. If you will be very explicit about what you need to prove, then you will already have 80% of the proof.

Ittay Weiss
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  • thanks. One question though, would it make a difference to the proof if instead I chose to use the definition Z(G)={g∈G∣xg=gx for all x∈G} ? – Sai82 May 03 '14 at 01:48
  • no, it would make no significant difference since $gx=xg$ if, and only if, $g^{-1}xg=x$. What is wrong with your answer is that it is unclear (perhaps to yourself) what exactly it is you assume at each step in the proof and what it is you are trying to show. Hence my advice: be explicit about it! – Ittay Weiss May 03 '14 at 01:58