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I need help with this advanced algebra problem.

Let $G$ be a group. We call the set $C(G)= \{a \in G : ab=ba, \forall b \in G\}$ the center of $G$. Prove that:

  (a)   $C(G)$ is normal subgroup of $G$.

  (b)   $C(G)=G$ if and only if $G$ is abelian group.

  (c)   If $a$ is the only element in $G$, then $a\in C(G)$.

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    @KCd: In fairness, they could just be problems from a book (and so not homework per se) However, the "*need*" a solution is intriguing...2012-06-07

4 Answers 4

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@Adam, first of all do not write $C(G)$ but $Z(G)$ for the center. This is common in group theory. Secondly, what you mean is if $a$ is the only element of order 2 (and not "second order").
If that is the case then observe that for every element $g \in G$, the conjugate $g^{-1}ag$ also has order 2. Hence $a = g^{-1}ag$ for all $g \in G$, that is $a \in Z(G)$.

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Hints

  1. $xax^{-1}=a$ for $a\in C(G)$.
  2. If $b\in G\setminus C(G)$. Do you have $x b=bx$ for each $x\in G$?
  3. What is your favorite group with one element?
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(a) Let $g \in G$, suppose $a \in C(G)$, $gag^{-1} = agg^{-1} = a$.

(b) If $G = C(G)$, then $a \in G$ implies $a \in C(G)$ so for any $b \in G$, $ab = ba$. Suppose $G$ is abelian, then clearly $ab = ba$ for all $a$.

(c) If $a$ is the only only element of $G$, then $a$ must be the identity. Clearly the identity element is in the center.

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    You should also prove that it is a subgroup...2012-06-06
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  • If $C(G)$ is the subgroup of elements that commute in $G$, don't they satisfy the axioms for a normal subgroup?
  • In an abelian group, all the elements commute.
  • If $a\in C(G)$, it commutes with all the other elements. If it is the only element, does it commute with all the "other" elements?
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    @Adam: Use `*text*` for italics, not math mode - I have corrected your comment above. [See here for more formatting commands](http://meta.stackoverflow.com/editing-help).2012-06-06