# The isomorphism theorems in group theory

These are some of the most important theorems in group theory. In this post, we’ll provide proofs, and explain some consequences of these theorems.

Theorem (first isomorphism theorem): Suppose $f:G\to H$ is a homomorphism of groups. Then: $\ker f$ is a normal subgroup of $G$$\mathrm{im}(f)=f(G)$ is a subgroup of $H$, and there is an isomorphism $\mathrm{im}(f)\cong G/\ker f$If $f$ is a surjection, then $H\cong G/\ker f$.

The proof is uses a lot of concepts, so tread carefully.

Proof: We’ve already proven the first two statements. We’ll now prove the last one. Define the map $\phi:G/\ker f\to\mathrm{im}(f)$ as $\phi:g\ker f\mapsto f(g)$. This is well defined, as we saw earlier. It’s a homomorphism. We want to prove that it’s bijective, i.e., surjective and injective. It’s obviously surjective by definition.

We need to prove that it’s injective, i.e., if $\phi(g\ker f)=\phi(h\ker f)$, then $g\ker f=h\ker f$. Recall that this is equivalent to proving that the kernel is trivial, i.e., all we need to prove is that if $\phi(g\ker f)=e$, then $g\ker f=\ker f$. Well, if $\phi(g\ker f)=e$, then $f(g)=e$. So $g\in\ker f$; but this implies that $g\ker f=\ker f$, so it is indeed injective! Thus it’s a bijective homomorphism, i.e., an isomorphism!

If the map is surjective, then $\mathrm{im}(f)=H$, so $H\cong G/\ker f$. QED.

Here is a cool application. Recall that $\mathrm{GL}_n(\mathbf{R})$ denotes the set of invertible $n\times n$ matrices of real numbers. Recall that for a $2\times 2$ matrix you can define the determinant function, which takes $\begin{smallmatrix} a&b\\ c&d \end{smallmatrix}$ to $ad-bc$. Analogously, the determinant of a $3\times 3$ matrix $\begin{smallmatrix} a&b&c\\ d&e&f\\g&h&k \end{smallmatrix}$ is $a(ek-fh)-b(dk-fg)+c(dh-eg)$. All of this can be generalized to $n\times n$ matrices. What you should notice is that the determinant of an invertible matrix of real numbers is never zero, and is always a real number.

The group of all invertible matrices whose determinant is $1$ is defined to be $\mathrm{SL}_n(\mathbf{R})$. This is a subgroup of $\mathrm{GL}_n(\mathbf{R})$. Now, the determinant function is a homomorphism $\mathrm{GL}_n(\mathbf{R})\to\mathbf{R}^\times$, where $\mathbf{R}^\times$ denotes the real numbers with the number zero removed, where the operation is multiplication. Well, the identity is $1$, so the kernel of this homomorphism consists of those invertible matrices whose determinant is $1$, and this is nothing but $\mathrm{SL}_n(\mathbf{R})$! So $\mathbf{R}^\times\cong\mathrm{GL}_n(\mathbf{R})/\mathrm{SL}_n(\mathbf{R})$ by the first isomorphism theorem.

Now onto the second isomorphism theorem.

Theorem (second isomorphism theorem): Suppose $H\leq G$, and let $K$ be a normal subgroup of $G$Define $HK=\{hk|h\in H,k\in K\}$. Then $HK\leq G$$H\cap K\leq H$ is a normal subgroup, and there is an isomorphism: $(HK)/K\cong H/(H\cap K)$.

Proof: For the first part: all we need to prove that it’s closed under multiplication and inverses. For multiplication: Suppose $g,j\in HK$. Then $g=h_1k_1$ and $j=h_2k_2$. So $gj=(h_1k_1)(h_2k_2)=h_1(k_1h_2)k_2$. Because $K$ is normal, you have $h_2^{-1}k_1h_2=m$ for some $m\in K$. So $k_1h_2=h_2m$ by multiplication on the left. So $gj=(h_1h_2)(mk_2)$, and this is an element of $HK$ because $h_1h_2\in H$ and $mk_2\in K$. For inverses: suppose $g\in HK$. Then $g=h_1k_1$. We want to show that $g^{-1}\in HK$. Well, $g^{-1}=k_1^{-1}h_1^{-1}$. For convenience, write $k_2=k_1^{-1}$ and $h_2=h_1^{-1}$. Then because $K$ is normal, $h_2^{-1}k_2h_2=m$ for some $m\in K$, and thus $k_2h_2=h_2m$. So $g^{-1}=h_2m$, and it’s therefore in $HK$.

For the second part: we want to prove that if $h\in H$, then $h^{-1}(H\cap K)h\subseteq (H\cap K)$. Suppose $g\in H\cap K$. Then $h^{-1}gh$ is in $H$, because $g\in H$ and $h\in H$ and $H$ is a group, i.e., is closed under multiplication. Because $K$ is normal, and $g\in K$ as well, by definition, $h^{-1}gh\in K$. Thus for any $g\in H\cap K$, we have $h^{-1}gh\in H$ and $h^{-1}gh\in K$, i.e., $h^{-1}gh\in (H\cap K)$.

For the last part: consider the map $H\to HK$ defined by $h\mapsto he$. Then you can compose with the map $HK\to (HK)/K$, which takes $a\mapsto aK$. So the total map $H\to HK\to (HK)/K$ takes $h\mapsto hK$. Denote this composition by $\phi$. OK, so now let’s consider some random $x\in (HK)/K$. Then $x=(hk)K=(hK)(kK)=(hK)(K)=hK$, so what we observe is that $\phi$ is surjective. Now, let’s ask: what is $\ker \phi$? Well, $\ker\phi=\{h\in H|\phi(h)=K\}$, which is $\{h\in H|hK=K\}=\{h\in H|h\in K\}$. But this is the definition of $H\cap K$. So $\ker\phi\cong H\cap K$, which is a group because the intersection of subgroups is itself a subgroup. Now we use the first isomorphism theorem, which says that $(HK)/K$ is isomorphic to $H/\ker\phi=H/(H\cap K)$. QED.

The very last isomorphism theorem is as follows.

Theorem (third isomorphism theorem): Suppose $K\leq H\leq G$ are normal subgroups. Then $(G/K)/(H/K)$ is isomorphic to $G/H$.

Proof: Recall the natural map $G\to G/H$ defined by $g\mapsto gH$. The kernel is $H$, and it contains $K$. Define $f:G/K\to G/H$ by $f:gK\mapsto gH$. Obviously this map is surjective. Now, its kernel is $H/K$ because if $gK\in\ker f$, then $gH=H$ and $g\in H$. So $\ker f$ consists of cosets $hK$ for $h\in H$. But this is just $H/K$. Now the first isomorphism theorem tells us that $(G/K)/\ker f\cong G/H$. But $\ker f=H/K$, so $(G/K)/(H/K)=G/H$. QED.

This is kind of like cancellation. For example, this tells us that $(\mathbf{Z}/n\mathbf{Z})/(n\mathbf{Z}/m\mathbf{Z})\cong\mathbf{Z}/n\mathbf{Z}$ if $m\mathbf{Z}\subseteq n\mathbf{Z}\subseteq \mathbf{Z}$. I guess we’ll end here for now; next time, we’ll go on to the Sylow theorems in group theory.

Have fun,
SKD