## Remarks on the isomorphism theorems

In the second theorem, we have $AB/B$ isomorphic to $A/A\cap B$ assuming $A$ is contained in $N_{G}(B)$. $B$ is normal in $AB$ because $abB=aB=Ba$ since $aBa^{-1}=B$.

This isomorphism comes from the maps \(A\to AB\to AB/B\) where the first is inclusion and the second is projection.

THe kernel of the composition is $A\cap B$, and the composition is surjective because every coset $abB$ is of the form $aB$. This proves that $A\cap B$ is normal in $A$, since it’s the kernel of a homomorphism.

In the Third theorem, we want to understand $(G/H)/(K/H)=G/K$ if $H$ and $K$ are normal in $G$ and $H$ is normal in $K$. The key is to use the map \(G/H\to G/K\) given by $aH\mapsto aK$. This is well defined since $K$ contains $H$, and surjective since $gK$ is the image of $gH$. The kernel of this map is the set of $gH$ such that $g\in K$, which is exactly $K/H$.

## Some applications of the isomorphism theorems

**Proposition:** Every subgroup of a cyclic group of order $n$ is cyclic of order a divisor of $n$, and there is a unique such subgroup for every divisor $d$ of $n$.

Suppose $G$ is cyclic of order $n$. This means that $G$ is isomorphic to the quotient group $\Zn{n}$. If $H$ is a subgroup of $G$, then $H$ must be of the form $\tilde{H}/n\Z$ where $\tilde{H}$ is a subgroup of $\Z$ containing $n\Z$. This means that $H=d\Z$ where $d$ is a divisor of $n$, and $H/n\Z=d\Z/n\Z$ is isomorphic to $\Zn{n/d}$.

Conversely, suppose $d$ divides $n$. Then $d\Z$ contains $n\Z$, so $d\Z/n\Z$ is a subgroup of $\Zn{n}$ of order $n/d$.

Thus every subgroup of $\Zn{n}$ is cyclic of order dividing $n$, and there is a subgroup of each such order.

Finally suppose $H_1$ and $H_2$ are two subgroups of order $d$ dividing $n$. Then \({\tilde{H}}_{1}\) and $\tilde{H}_{2}$ are subgroups of $\Z$ of index $n/d$; since there is only one such subgroup, they are the same. Therefore there is a unique subgroup of $G$ of each order $d$ dividing $n$.

Note that, in the proof above, we never mention any elements of the group.

## The commutator subgroup

Suppose that $G$ is an arbitrary group. When is there a homomorphism from $G$ to an abelian group?

Suppose $A$ is abelian and $f:G\to A$ is a surjective homomorphism. Then every element of $G$ of the form $xyx^{-1}y^{-1}$, where $x$ and $y$ are in $G$, is in the kernel of $f$.

Therefore the subgroup of $G$ generated by these elements is in the kernel of $f$.

**Note:** Not every product of commutators is a commutator, so you really need $[G,G]$ to be the subgroup *generated* by the commutators. In fact the matrix $-I_{2}$ in $\SL_{2}(\R)$ is not a commutator.

**Definition:** Let $G$ be a group. If $x,y\in G$, let $[x,y]=xyx^{-1}y^{-1}$. This is called the commutator of $x$ and $y$. Let $[G,G]$ be the subgroup of $G$ generated by all such commutators. This is called the commutator subgroup of $G$.

Example: in the Dihedral group $D_{2n}$, the commutator subgroup is generated by $r^{2}$, so depending on whether $n$ is even or odd the commutator subroup is $R$ or its subgroup of index $2$.

**Proposition:** Let $G$ be a group. Then the commutator subgroup of $G$ is normal. If $f:G\to A$ is a homomorphism where $A$ is abelian, then there is a commutative diagram

where $\pi:G\to [G,G]$ is the quotient map. In other words, every homomorphism from $G$ to an abelian group $A$ “comes from” a homomorphism from $G/[G,G]$ to $A$.

This is the first isomorphism theorem once we observe that $G/[G,G]$ is abelian and $[G,G]$ is in the kernel of every map to an abelian group.

The commutator subgroup of $\GL_{2}(\R)$ is $\SL_{2}(R)$. (This is proved by somewhat painful calculations.)