# Abstract Nonsense

## Review of Group Theory: The Center of a Group

Point of post: In this post we take a brief break from our discussion of group actions to address the notion of the center of a group.

Motivation

Just as was the case for the center of an algebra it is often fruitful to consider the center of a group. In essence, the center of a group is merely the set of all elements of the group which commute with all other elements of the group. We shall use the concepts in this post in this and the next post to conclude that for any group of order $p^2$ with $p$ prime must be abelian.

The Center of a Group

Let $G$ be a group, we define the center of $G$ to be

$\mathcal{Z}\left(G\right)=\left\{g\in G:gg'=g'g,\;\;\text{for every }g'\in G\right\}$

Our first theorem says that the center of a group is a normal subgroup of a group:

Theorem: Let $G$ be a group, then $\mathcal{Z}\left(G\right)\trianglelefteq G$. Moreover, $G/\mathcal{Z}(G)\cong\text{ Inn}(G)\leqslant \text{Aut}(G)$ (where $\text{Inn}(G)$ is the set of inner automorphisms).

Proof: Recall that $\Phi:G\to \text{Aut}(G):g\mapsto i_g$ is a homomorphism. Note then that

\displaystyle \begin{aligned}\ker\Phi &= \left\{g\in G:i_g=\text{id}_G\right\}\\ &= \left\{g\in G:g^{-1}hg=h\;\;\text{for all }h\in G\right\}\\ &=\left\{g\in G:hg=gh\;\;\text{for all }\in G\right\}\\ &=\mathcal{Z}\left(G\right)\end{aligned}

from where the conclusion follows from our earlier characterization of normality. The fact that $G/\mathcal{Z}(G)\cong\text{ Inn}(G)$ now follows immediately from the First Isomorphism Theorem.$\blacksquare$.

An important theorem involving the center of a group is the following

Theorem: Let $G$ be a group. Then, if $G/\mathcal{Z}(G)$ is cyclic then $G$ is abelian.

Proof: Let $\mathcal{Z}(G)=\left\langle g\mathcal{Z}\left(G\right)\right\rangle$. Then, for any $a,b\in G$ we have that there exists $n,m\in\mathbb{Z}$ such that $a\mathcal{Z}(G)=g^n\mathcal{Z}(G)$ and $b\mathcal{Z}(G)=g^m\mathcal{Z}(G)$. Consequently there exists $z_1,z_2\in\mathcal{Z}(G)$ such that $a=g^nz_1$ and $b=g^mz_2$. Thus,

\begin{aligned}ab &=\left(g^nz_1\right)\left(g^mz_2\right)\\ &=g^nz_1z_2g^m\\ &=(z_1z_2)g^ng^m\\ &=(z_1z_2)g^mg^n\\ &=g^m(z_1z_2)g^n\\ &=\left(g^mz_2\right)\left(g^nz_1\right)\\ &=ba\end{aligned}

the conclusion follows. $\blacksquare$

The reason why this is slightly important is that it well let us conclude (later) that the only groups up to isomorphism of order $p^2$ where $p$ is prime are $C_{p^2}$ or $C_p\oplus C_p$ where $C_k$ is a symbol  for the generic cyclic group of order $k$ and $\oplus$ is the direct product of groups which we have yet to discuss.

References:

1.  Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200

January 5, 2011 -

1. […] reasons) and is called the conjugacy class of . We note the trivial fact that (where is the center of ) if and only if […]

Pingback by Review of Group Theory: Group Actions (Pt. IV Conjugation and the Class Equation) « Abstract Nonsense | January 6, 2011 | Reply

2. […] reasons) and is called the conjugacy class of . We note the trivial fact that (where is the center of ) if and only if […]

Pingback by Review of Group Theory: Group Actions (Pt. IV Conjugation and the Class Equation) « Abstract Nonsense | January 6, 2011 | Reply

3. […] Assume that then and thus (from previous theorem) we may conclude that is cyclic. But, from an earlier theorem we may then conclude that is abelian and thus which is a contradiction. Thus, it follows that […]

Pingback by Review of Group Theory: Group Actions (Pt. IV Conjugation and the Class Equation Pt. II) « Abstract Nonsense | January 6, 2011 | Reply

4. […] is that is a normal subgroup of . Indeed it clearly suffices to check that for every . Recall the inner automorphism and the fact that it is, surprisingly, an automorphism. Thus, one sees that and since were […]

Pingback by Review of Group Theory: The Commutator Subgroup and the Abelianization of a Group « Abstract Nonsense | February 27, 2011 | Reply

5. […] further and show that for any finite group and any one has that (where, as usual, denotes the center of the group). This proof is quite long, but the result is interesting enough to merit the effort. […]

Pingback by Representation Theory: The Dimension Theorem (Strong Version) « Abstract Nonsense | March 7, 2011 | Reply

6. […] We use the fact that since that is inevitably cyclic and so by a common theorem we must have that is abelian. Then, by the structure theorem we may conclude that either or […]

Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) ( January 2003)) « Abstract Nonsense | May 1, 2011 | Reply

7. […] get an idea of these subrings let’s define the analogous idea of the center of a group for a ring and show it’s a subring. Namely, if is a ring then we define the center of to […]

Pingback by Subrings « Abstract Nonsense | June 15, 2011 | Reply