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 $aB{a}^{-1}=B$.

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

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

Thus every subgroup of $\mathbb{Z}/n\mathbb{Z}$ 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 $\mathbb{Z}$ 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 $xy{x}^{-1}{y}^{-1}$, 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 ${\mathrm{SL}}_2(\mathbb{R})$ is not a commutator.

Definition: Let $G$ be a group. If $x,y\in G$, let $[x,y]=xy{x}^{-1}{y}^{-1}$. 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

$$G\pi fAG/[G,G]\bar{f}$$

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 ${\mathrm{GL}}_2(\mathbb{R})$ is ${\mathrm{SL}}_2(R)$. (This is proved by somewhat painful calculations.)

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