# 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

**Cite as:**Group Isomorphism Theorems.

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