# Group Isomorphism Theorems

In group theory, two groups are said to be **isomorphic** if there exists a bijective homomorphism (also called an isomorphism) between them. An isomorphism between two groups $G_1$ and $G_2$ means (informally) that $G_1$ and $G_2$ are the same group, written in two different ways.

Many groups that come from quotient constructions are isomorphic to groups that are constructed in a more direct and simple way. There are three standard isomorphism theorems that are often useful to prove facts about quotient groups and their subgroups.

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## First Isomorphism Theorem

This theorem is the most commonly used of the three. Given a homomorphism between two groups, the **first isomorphism theorem** gives a construction of an induced isomorphism between two related groups.

Let $G$ and $H$ be two groups and let $\phi\colon G\to H$ be a group homomorphism. Then the kernel $\textrm{ker}(\phi)$ is a normal subgroup of $G,$ and

$G/\textrm{ker}(\phi) \simeq \textrm{Im}(\phi).$

Here $\textrm{Im}(\phi)$ is the image of the homomorphism $\phi,$ which is a subgroup of $H.$ The isomorphism $\tilde{\phi}$ is completely natural: letting $N = \textrm{ker}(\phi),$ $\tilde{\phi}$ sends an element $gN$ of the quotient $G/N$ to the image of $g$ under $\phi.$ That is, $\tilde{\phi}(gN) = \phi(g).$

Let ${\mathbb C}^*$ be the group of nonzero complex numbers under multiplication. Consider the homomorphism $f \colon {\mathbb C}^* \to {\mathbb C}^*$ given by $f(z) = z^2.$ Then the first isomorphism theorem says that ${\mathbb C}^*/\text{ker}(f) \simeq \text{im}(f).$

Now $\text{ker}(f) = \{ \pm 1\},$ and $\text{im}(f) = {\mathbb C}^*$ (that is, $f$ is onto), so the conclusion is that ${\mathbb C}^*/\{\pm 1\} \simeq {\mathbb C}^*,$ i.e. ${\mathbb C}^*$ is isomorphic to a nontrivial quotient of itself.

What is wrong with this argument?

## Second Isomorphism Theorem

The **second isomorphism theorem** relates two quotient groups involving products and intersections of subgroups.

Let $G$ be a group, let $H$ be a subgroup, and let $N$ be a normal subgroup. Then $HN = \{hn : h \in H, n \in N\}$ is a subgroup of $G,$ and

$HN/N \simeq H/(H \cap N).$

Let $S_3$ be the symmetric group on three letters. Let $H$ be the subgroup generated by the transposition $(12),$ and let $N$ be the subgroup generated by the transposition $(13).$

Then the second isomorphism theorem gives an isomorphism $HN/N \simeq H/(H \cap N),$ where $HN = \{ hn : h \in H, n \in N\}.$

Now $H \cap N = \{1\},$ so the right side has order $2.$ So the left side has order $2.$ Now $|N|=2$ and $|HN/N|=2,$ so $|HN|=4.$ But Lagrange's theorem says that $|HN|$ must divide $|S_3|,$ which is $6.$

What has gone wrong with this argument?

**I.** As defined above, $HN$ is not a subgroup of $S_3,$ so Lagrange's theorem does not apply.

**II.** The order of a quotient $G/K$ of finite groups is not always equal to $|G|/|K|.$

**III.** $N$ is not normal, so the second isomorphism theorem does not apply.

## Third Isomorphism Theorem

The **third isomorphism theorem** is extremely useful in analyzing the normal subgroups of a quotient group.

Let $G$ be a group, and $N$ a normal subgroup. Then

(1) There is a natural one-to-one correspondence between subgroups of $G/N$ and subgroups of $G$ containing $N.$ The correspondence matches a subgroup $K$ of $G$ containing $N$ with the subgroup $K/N$ of $G/N.$

(2) This correspondence preserves normality: the normal subgroups of $G/N$ correspond to the normal subgroups of $G$ containing $N.$ That is, a subgroup $K$ of $G$ containing $N$ is normal in $G$ if and only if $K/N$ is normal in $G/N.$

(3) The correspondence preserves quotients: if $K$ is a normal subgroup of $G$ containing $N,$ then $(G/N)/(K/N) \simeq G/K.$

## Examples

Consider the group $GL_2({\mathbb R})$ of invertible matrices. The determinant is a homomorphism from $GL_2({\mathbb R})$ to the multiplicative group ${\mathbb R}^*$ of nonzero real numbers. It is surjective as well (for any $a \ne 0,$ the determinant of $\small \begin{pmatrix} a&0 \\ 0&1 \end{pmatrix}$ is $a$). Then the first isomorphism theorem says that $GL_2({\mathbb R})/N \simeq {\mathbb R}^*,$ where $N$ is the kernel of the determinant homomorphism. Here, $N$ is simply the set of matrices which the homomorphism sends to the identity $1$ of ${\mathbb R}^*,$ i.e. the subgroup of matrices with determinant $1.$ This is called $SL_2({\mathbb R}).$ So $GL_2({\mathbb R})/SL_2({\mathbb R}) \simeq {\mathbb R}^*$ by the first isomorphism theorem.

A nontrivial consequence of the third isomorphism theorem is that any subgroup of $GL_2({\mathbb R})$ containing $SL_2({\mathbb R})$ is normal, because the quotient group is abelian, so all its subgroups are normal. The third isomorphism theorem also gives a full description of the subgroups of $GL_2({\mathbb R})$ containing $SL_2({\mathbb R})$; they consist of the matrices whose determinants lie in a subgroup of ${\mathbb R}^*.$

Suppose $x,y$ are positive integers with $x|y.$ Then $y{\mathbb Z} \subseteq x{\mathbb Z}.$ Find the order of the quotient $x{\mathbb Z}/y{\mathbb Z}.$

This is an application of the third isomorphism theorem. Let $G = {\mathbb Z}, H = x{\mathbb Z}, N = y{\mathbb Z}.$ Then $G/H \simeq (G/N)/(H/N).$ Now $|G/H| = x$ and $|G/N| = y,$ so $|H/N| = y/x.$ $_\square$

Prove that $\text{lcm}(a,b) \text{gcd}(a,b) = ab$ using the second isomorphism theorem and the previous example.

Let $G= {\mathbb Z}, H = a{\mathbb Z}, N = b{\mathbb Z}.$ The group law is addition, so the second isomorphism theorem becomes $\frac{a{\mathbb Z} + b{\mathbb Z}}{b{\mathbb Z}} \simeq \frac{a{\mathbb Z}}{a{\mathbb Z} \cap b{\mathbb Z}}.$ Now $a{\mathbb Z} \cap b{\mathbb Z} = \text{lcm}(a,b){\mathbb Z}$, since it is the set of multiples of both $a$ and $b.$

On the other hand, $a{\mathbb Z} + b{\mathbb Z}$ is the set of all integers of the form $ax+by,$ $x,y \in {\mathbb Z}.$ A well-known result in elementary number theory known as Bezout's identity says that this set consists of the multiples of $\text{gcd}(a,b).$ So $\frac{\text{gcd}(a,b){\mathbb Z}}{b{\mathbb Z}} \simeq \frac{a{\mathbb Z}}{\text{lcm}(a,b){\mathbb Z}}.$ Taking the orders of both sides and using the previous example gives $\frac{b}{\text{gcd}(a,b)} = \frac{\text{lcm}(a,b)}{a},$ and the result follows. $_\square$

## Proofs

**Proof of the first isomorphism theorem:**

To prove the first theorem, we first need to make sure that $\operatorname{ker} \phi$ is a normal subgroup (where $\operatorname{ker} \phi$ is the kernel of the homomorphism $\phi$, the set of all elements that get mapped to the identity element of the target group $H$). This means we need to prove that for any homomorphism, the identity element is in the kernel, the inverse of elements in the kernel are also in the kernel, the product of two elements in the kernel is also in the kernel, and for any

$g \in G , n \in \operatorname{ker} \phi \longrightarrow g^{-1} n g \in \operatorname{ker} \phi$.

To see that the identity element of $G$ is an element of $\operatorname{ker} \phi$, consider $\phi(e_G g) , g \in G$, where $e_G$ denotes the identity element of $G$. On one hand:

$\phi(e_G g) = \phi(e_G) \phi(g)$,

but on the other hand,

$\phi(e_G g) = \phi(g)$,

so we must have

$\phi(e_G) \phi(g) = \phi(g) \forall g \in G \Longrightarrow \phi(e_G) = e_H \Longrightarrow e_G \in \operatorname{ker} \phi$.

We similarly prove the existence of inverse in $\operatorname{ker} \phi$: if $n \in \operatorname{ker} \phi$, then

$\phi(n^{-1}) \phi(n) = \phi(n^{-1} n) = \phi(e_G) = e_H$,

$\phi(n^{-1}) \phi(n) = \phi(n^{-1})$, so

$\phi(n^{-1}) = e_H \forall n \in \operatorname{ker} \phi \Longrightarrow n^{-1} \in \operatorname{ker} \phi$.

Closure is easy to prove:

$\phi(n_1 n_2) = \phi(n_1) \phi(n_2) = e_H \Longrightarrow (n_1)(n_2) \in \operatorname{ker} \phi \forall n_1 , n_2 \in \operatorname{ker} \phi$.

To prove that $\operatorname{ker} \phi$ is normal within $G$, note that for any $n \in \operatorname{ker} \phi$ and any $g \in G$,

$\phi( g^{-1} n g) = \phi(g^{-1}) \phi(n) \phi(g) = \phi(g^{-1}) \phi(g) = \phi(g^{-1} g) = \phi(e_G) = e_H$,

so $g^{-1} n g \in \operatorname{ker} \phi \forall g \in G , n \in \operatorname{ker} \phi$.

The proof that $\operatorname{Im} \phi$ is also a group can be accomplished by similar means. For simplicity's sake, let's refer to $\operatorname{ker} \phi$ as $K$ and $\operatorname{Im} \phi$ as $I$. The quotient group

$G/K$

is defined as the group of cosets $g K, g \in G$, with multiplication given by $(a K)(b K) := (ab) K$. To prove the isomorphism between this group and $I$, consider the homomorphism

$\widetilde{\phi} \colon G/K \longrightarrow H , \widetilde{\phi} (a K) = \phi(a) \forall a \in G$.

By the definition of the image of a function, any function is surjective, or onto, into its image, so we just need to prove that $\widetilde{\phi}$ is injective, or one-to-one:

$\widetilde{\phi}(a K) = \widetilde{\phi}(b K) \Longrightarrow a K = b K$.

Start with the equation

$\phi(a) = \phi(b)$.

Left-multiply both sides by $\phi(b^{-1})$ to get:

$\phi(b^{-1} a) = e_H \Longrightarrow b^{-1} a \in K$.

This means that

$b^{-1} a = f$ for some $f \in K$, and

$a = b f \in b K$, and $b = a f^{-1} \in a K$.

Because of the closure of $K$ under multiplication, every element of the form $a f' , f' \in K$ will also be in $b K$. But these are precisely all the elements of $a K$. So $a K \subset b K$. And the second relationship shows us that $b K \subset a K$. So we must have $b K = a K$. This completes the proof of the first isomorphism theorem.

**Proof of the second isomorphism theorem:**

This proof relies on the first isomorphism theorem. It is easy to see that $N$ is normal within $HN$ and $H \cap N$ is normal within $H$. Since both sides of the isomorphism are quotient groups, it would be nice if we could construct homomorphisms $\phi_1 \colon HN \longrightarrow K_1$ with $\operatorname{ker} \phi_1 = N$, and $\phi_2 \colon H \longrightarrow K_2$ with $\operatorname{ker} \phi_2 = H \cap N$, then prove that $\operatorname{Im} \phi_1 \simeq \operatorname{Im} \phi_2$. Fortunately, one of the many equivalent definitions of a normal subgroup is that there exists a homomorphism whose kernel is that subgroup. Let $\phi_1$ be the homomorphism from $HN$ to some group $K$ which satisfies this property for $N$. All the elements of $HN$ can be expressed as $hn, h \in H , n \in N$. Notice that:

$\phi_1 (hn) = \phi_1 (h) \phi_1 (n) = \phi_1 (h)$, and this will be different from the identity element if and only if $h \notin N$. This means

$\operatorname{Im} \phi_1 = \{ \phi_1 (h) | h \in HN \} = \{ \phi_1 (h) | h \in H \}$.

Now consider the same homomorphism applied to the subgroup $H$ of $HN$. Its kernel will be all the elements of $H$ which are also elements of $N$, or $H \cap N$. Its image when it's applied to this group is defined as $\operatorname{Im} \phi_1 |_H = \{ \phi_1 (h) | h \in H\}$, which is equal to the image of $\phi_1$ when it's applied to $HN$. So the two quotient groups

$HN/N$ and $H/(H \cap N)$

are both isomorphic to the same group, $\operatorname{Im} \phi_1$. Therefore they are isomorphic to one another.

**Cite as:**Group Isomorphism Theorems.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/group-isomorphism-theorems/