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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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Is this homework? – Eugene Jun 6 '12 at 19:00
This question doesn't show much effort on the part of the questioner to solve the problem – Ben Millwood Jun 6 '12 at 19:06
I agree this looks like homework and does not show much effort. – Nate Iverson Jun 6 '12 at 19:36
The centre is usually denoted by $Z(G)$, the $Z$ standing for zentrum (which is the German for centre - pre-war maths was all German...) – user1729 Jun 6 '12 at 20:58
I am curious: since this is not homework, in exactly what kind of context did such a sequence of problems arise for you? The first part is a routine result that is in most abstract algebra books (comparable to discussions of $dy/dx = ay$ in calculus books), and part c seems very strange as anything other than homework; I can't imagine how else part c could be a task. – KCd Jun 7 '12 at 1:59
up vote 2 down vote accepted

(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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Fixed it, but does it really matter. – William Jun 6 '12 at 19:05
You should also prove that it is a subgroup... – user1729 Jun 6 '12 at 20:56


  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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  • 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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How can I prove the third statement if a is the only element of second order in the group? – Adam Jun 6 '12 at 19:35
If $a$ is the only element of a group, then $a$ is the identity element $a=1$. Clearly, the identity commutes with all other elements, $1\in G, 1.b=b.1 \forall b\in G$ – zzzzzzzzzzz Jun 6 '12 at 19:38
It's clear. Thank you. Is the same statement true, if a is the only element of second order in the group? – Adam Jun 6 '12 at 19:41
Yes. if $a^2=1$, then obviously $1.a=a.1$ and $a*a=1=a*a$, the element commutes with itself. – zzzzzzzzzzz Jun 6 '12 at 19:45
@Adam: Use *text* for italics, not math mode - I have corrected your comment above. See here for more formatting commands. – Zev Chonoles Jun 6 '12 at 19:48

@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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