Examples of non-obvious isomorphisms following from the first isomorphism theorem

Question

I am learning the first isomorphism theorem, and I am working with some isomorphisms to practice for my upcoming test. I know some of the basic ones like:

• $\mathbb{R}/\mathbb{Z} \cong \mathcal{C}$, where $\mathcal{C}$ is the unit circle in the complex plane, under the isomorphism $$x+\mathbb{Z}\mapsto e^{2\pi x i}$$

• $\dfrac{\mathbb Z \times \mathbb Z}{\langle (m,n)\rangle}\cong\mathbb Z$, where $m,n$ are integers.

• $\dfrac{\mathbb R^\star}{\{1, -1\}} \cong \mathbb R^+$.

I would like to see more examples of such isomorphisms, intended both as a reference and to help me study for the test! Thank you.

Answer 1

First Isomorphism Theorem examples

First Isomorphism Theorem (FIT) applies in different contexts: Groups, Rings, Vector Spaces, Lie Algebras and other structures.

As follows, examples for the first three.

Three Group Isomorphisms

Let us consider $GL_2(\Bbb F_3)$: the group of $2\times 2$ invertible matrices with coefficients in $\Bbb F_3$ which is the field with three elements. This is a group wrt the matrix multiplication. Consider now the linear map $\det:GL_2(\Bbb F_3)\to\Bbb F_3^{\times}$. It's clearly surjective, and its kernel are all the matrices with determinant equal to $1$, which form a normal subgroup denoted by $SL_2(\Bbb F_3)$.

Thus from FIT it follows that $$ \color{#c01}{\frac{GL_2(\Bbb F_3)}{SL_2(\Bbb F_3)}\simeq\Bbb F_3^{\times}}\;. $$

Another interesting example which descend from FIT is the following (but proving it is not immediate): let $\Bbb F$ be a field, and $t$ a trascendental element over $\Bbb F$. It can be proven that $$ \color{#b49}{\operatorname{Aut}(\Bbb F(t)|\Bbb F)\simeq GL_2(\Bbb F)/D_2(\Bbb F)} $$ where $D_2(\Bbb F):=\{a\Bbb I_2\;:\;a\in\Bbb F\}$ are the $2\times2$ scalar matrices.

A last interesting example with groups: consider the additive group $H:=(\Bbb Z/m\Bbb Z,+)$. Then $$ \color{#d78}{\operatorname{Aut}(H)\simeq U(\Bbb Z/m\Bbb Z)} $$ where $U(\Bbb Z/m\Bbb Z)$ is the group of unit (i.e. the invertible elements) of the ring $\Bbb Z/m\Bbb Z$. Try to figure out the morphism between the two groups and check it is bijective.

A Ring Isomorphism

Consider a field $\Bbb K$, $\operatorname{char}(\Bbb K)=0$ and $\overline {\Bbb K}=\Bbb K$ (i.e. algebraically closed).

Consider the ring morphism given by $\varphi:\Bbb K[X,Y,Z]\to\Bbb K[X,Z]$ given by $X\mapsto X$, $Z\mapsto Z$ and $Y\mapsto XZ$.

Then $\varphi$ is clearly surjective, and the kernel is the ideal $(ZX-Y)$. Thus FIT gives $$ \color{#c33}{\frac{\Bbb K[X,Y,Z]}{(ZX-Y)}\simeq\Bbb K[X,Z]}\;. $$

A Vector Space Isomorphism

Consider the map $\phi:\Bbb R^4\to\Bbb R^2$ given by $\begin{bmatrix}x\\y\\z\\w \end{bmatrix}\mapsto\begin{bmatrix}x-z\\y+w \end{bmatrix}$.

$\phi$ is an $\Bbb R$-linear map, clearly surjective, in which the kernel is given by $ W:=\langle\begin{bmatrix}1\\0\\1\\0\end{bmatrix}, \begin{bmatrix}0\\1\\0\\-1 \end{bmatrix}\rangle $.

Hence, FIT gives $$ \color{#c03}{\Bbb R^4/W\simeq\Bbb R^2}\;\;. $$

Answer 2

Sometimes, instead of using the first isomorphism theorem as a tool to construct isomorphisms, it can be used as a tool to construct subgroups with certain properties. For example, consider the problem:

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Let $G$ be a finite group with subgroup $H$, $[G:H] = n$, then $H$ contains a normal subgroup of index $\leq n!$

Solution: $G$ acts on $G/H$ by $x(gH) = (xg)H$. This induces a homomorphism $\phi$ from $G$ to $\Sigma_{G/H}$. Suppose that $x$ is in the kernel of $\phi$. Then $xH = H$, so that $x \in H$. Thus the kernel $K$ of $\phi$ is contained inside $H$. By the first isomorphism theorem, $G/K$ is isomorphic to a subgroup of $\Sigma_{G/H}$, which necessarily has order less than $|\Sigma_{G/H}| = n!$.

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The purpose of this example is to demonstrate that you can produce a non-obvious subgroup by picking an appropriate homomorphism, looking at its kernel, and applying the first isomorphism theorem.

Answer 3

I really like isomorphisms involving matrix groups. For example, $SU(2)/\{\pm I\}\simeq SO(3)$ or $SL_2(\mathbb{F}_5)/\{\pm I\}\simeq A_5$. Here $SU(2)$ is the group of unitary $2\times 2$ matrices, $SO(3)$ is the group of real orthogonal $3\times 3$ matrices, $SL_2(\mathbb{F}_5)$ are matrices with coefficiens in the field with $5$ elements with $\det=1$, and $A_5$ is a group of even permutations of $5$ elements.

I am not sure if this is exactly what you are asking for, but I hope it helps a bit.

Answer 4

$(\mathbb{R^{+}},\bullet)$, the multiplicative group of positive real numbers, is isomorphic to $(\mathbb{R},+)$, the additive group of real numbers.

Consider logarithmic function $x \mapsto ln(x)$. Since $ln(ab)=ln(a)+ln(b)$ and it is totally defined on $\mathbb{R^{+}}$ and surjective, it is a homomorphism from $(\mathbb{R^{+}},\bullet)$ onto $(\mathbb{R},+)$. Its kernel is the set of the preimages of the identity element of $(\mathbb{R},+)$, so it is the trivial subgroup. Hence we have

$$(\mathbb{R^{+}},\bullet) / \{1\} \cong (\mathbb{R^{+}},\bullet) \cong (\mathbb{R},+)$$

Of course, it is also easy to see this isomorphism without using the first isomorphism theorem, but this is surely an example that illustrates this theorem. It involves two groups that should be well understood by those learning this theorem for the first time, and is non-obvious, since one of the groups is additive and another multiplicative.

Answer 5

$\mathbb{R}[i] \simeq \mathbb{C}$

The first isomorphism theorem is what allows you to say $\mathbb{R}[x]/(x^2+1) \simeq \mathbb{R}[i]$ which is just $\mathbb{C}$.

We identify the kernel of $\phi:p(x) \to p(x) \mod (x^2 + 1)$ with the ring setting $x^2 + 1 = 0$. This is a very powerful idea, leading to the notion of algebraic variety

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Brouwer Fixed Point Theorem (topology)

If you know some topology (e.g. Hatcher Chapter 1), a result from homotopy theory says that if $r:X \to A$ is a retraction, and $i:A \to A \subseteq X$ is an inclusion, then $i_\ast:\pi_1(A) \to \pi_1(X)$ is an injective map.

Let $X = \mathbb{D}$ the unit disk and $A=S^1 $ the boundary circle. Then $\pi_1(\mathbb{D}) = 0$ and $\pi_1(S^1) = \mathbb{Z}$. The inclusion of the boundary circle onto $S^1 \to \mathbb{D}$ induces a homomorphism between fundamental groups $i_{\ast}:\pi_1(S^1) \to \pi_1(\mathbb{D})$.

By the first isomorphism theorem we have $\boxed{\mathrm{ker\;}i_{\ast} \simeq \pi_1(S^1)/\pi_1(\mathbb{D})}$ unfortunately a closer look shows $i_\ast :\mathbb{Z} \to 0 $ so $\mathrm{ker\;}i_{\ast} \simeq \mathbb{Z}$ but if this homormophism were injective $\mathrm{ker\;}i_{\ast} \simeq 0$.

We conclude there exists no contraction, $\mathbb{D} \to S^1$.

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If $h:\mathbb{D} \to \mathbb{D}$ is continuous, then we could draw lines $x \to h(x)$ and see where it hits $S^1$. If $h(x) \neq x$, this would induce a retraction from $\mathbb{D} \to S^1$ which we showed doesn't exist. So we showed...

Brouwer Fixed Point Theorem Every continuous map $\mathbb{D} \to \mathbb{D}$ has a fixed point.

Can you see where I used the 1st isomorphism theorem?