Group Theory

Let $H$ be a normal subgroup of $G$. Then it can be verified that the cosets of $G$ relative to $H$ form a group. This group is called the quotient group or factor group of $G$ relative to $H$ and is denoted $G/H$.

It can be verified that the set of self-conjugate elements of $G$ forms an abelian group $Z$ which is called the center of $G$. Note the center consists of the elements of $G$ that commute with all the elements of $G$. Clearly the center is always a normal subgroup.

Theorem: A group $G$ of order $p^2$ where $p$ is prime is always abelian.

Proof: From a previous theorem, the number of invariant elements is a positive multiple of $p$, so the center has order $p$ or $p^2$. The latter case implies $G$ is abelian, so consider the case $\mid Z\mid =p$. Then $\mid G/Z\mid =p$ so $G/Z$ is cyclic, thus we may decompose $G$ into the cosets $Z,Zg,...,Z{g}^{p-1}$ for some $g\in G$. The product of any two elements $z_1{g}^{\lambda },z_2{g}^{\mu }$ is $z_1{z}_2{g}^{\lambda +\mu }=z_2{g}^{\mu }{z}_1{g}^{\lambda }$, thus $G$ is abelian and $\mid Z\mid =p^2$ in fact.

Define the commutator of two elements $g,h$ of a group $G$ by $u=g^{-1}{h}^{-1}gh$. We have $u=1$ if and only if $gh=hg$. In an abelian group, all commutators are equal to the identity. Consider the set of all commutators $\{u_1,...,u_m\}$ as $g,h$ run through all the elements of $G$. This set is not necessarily closed under the group operation. We define the commutator group $U$ to be the group generated by this set. If $U=G$ we say $G$ is a perfect group.

Theorem: The commutator group $U$ of a group $G$ is normal. $G/U$ is abelian. $U$ is contained in every normal subgroup that has an abelian quotient group.

Proof: Let $x\in G$. Then $x^{-1}{g}^{-1}{h}^{-1}ghx=a^{-1}{b}^{-1}ab$ where $a=x^{-1}gx,b=x^{-1}hx$, thus $U$ is normal.

Consider the commutator of two cosets $Ux,Uy$. We have $$(U{x}^{-1})(U{y}^{-1})(Ux)(Uy)=U{x}^{-1}{y}^{-1}xy=U$$ since $x^{-1}{y}^{-1}xy\in U$, hence $G/U$ is abelian.

Lastly if $R$ is any normal subgroup of $G$ with an abelian quotient group, then for any $x,y\in G$ we have $R{x}^{-1}{y}^{-1}xy=R$ since all commutators of $G/R$ must be equal to the identity, thus $R$ contains $x^{-1}{y}^{-1}xy$ hence $R\supset U$.

Theorem: If $A,B$ are normal subgroups of $G$ with only the identity element in common then every element of $A$ commutes with every element of $B$.

Proof: Consider $u=a^{-1}{b}^{-1}ab=(a^{-1}{b}^{-1}a)b=a^{-1}(b^{-1}ab)$ where $a\in A,b\in B$. Then since $A,B$ are normal, $a^{-1}ba\in B$ and $b^{-1}ab\in A$, thus $u\in A\cap B=\{1\}$, hence $a,b$ commute.