Notes:Group isomorphism theorems

Note: the bulk of the information on this page comes from:

• Books:Abstract Algebra - Pierre Antoine Grillet

Factorisation Theorem

Let $N$ be a normal subgroup of a group $(G,*)$. Let $\varphi:G\rightarrow H$ be any group homomorphism (so $H$ is also a group), whose kernel contains $N$, then $\varphi$ factors uniquely through $\pi:G\rightarrow G/N$.

$\xymatrix{ G\ar[d]_{\pi} \ar[r]^{\varphi} & H \\G/N \ar@{.>}[ur]_{\psi} & }$
This diagram commutes

That is to say that there exists a group homomorphism:

• $\psi:G/N\rightarrow H$

such that $\varphi=\psi\circ\pi$

In line with the factor (function) page, note that we require $\forall x,y\in G[(\pi(x)=\pi(y))\implies(\varphi(x)=\varphi(y))]$ which is not immediately obvious.

Additionally, notice that the induced function, $\psi:G/N\rightarrow H$ is unique, this is because the canonical projection map, $\pi:G\rightarrow G/N$ (which $\pi:g\mapsto gN$ where $gN$ denotes a coset of $N$) is surjective. This uniqueness claim is proved on the factor (function) page.

First Isomorphism Theorem

$\xymatrix{ A \ar[r]^{\varphi} \ar[d]_{\pi} & B \\ A/\text{Ker}(\varphi) \ar@{.>}[r]^-\theta & \text{Im}(\varphi) \ar@{^{(}->}[u]_{i} }$
This diagram commutes

If $\varphi:A\rightarrow B$ is any group homomorphism then:

• $A/\text{Ker}(\varphi)\cong\text{Im}(\varphi)$

In fact, there is a group isomorphism, $\theta:A/\text{Ker}(\varphi)\rightarrow\text{Im}(\varphi)$ which is unique, such that:

• $\varphi=i\circ\theta\circ\pi$ where $i:\text{Im}(\varphi)\rightarrow B$ is the inclusion map.

Notice that if we take $\bar{\varphi}:A\rightarrow\text{Im}(\varphi)$ being $\bar{\varphi}:a\mapsto\varphi(a)$ that this map has the same kernel as $\varphi$. This means we get a new diagram:

$\xymatrix{A \ar[r]^{\varphi} \ar[d]_{\pi} \ar@{-->}[dr]^(0.4){\bar{\varphi} } & B \\ A/\text{Ker}(\varphi) \ar@{.>}[r]^-\theta \ar@{.>}[ur]^(.3)\alpha & \text{Im}(\varphi) \ar@{^{(}->}[u]^{i} }$

Where: The dotted arrows are induced by the factorisation theorem, above. The dashed arrow is induced by $\varphi$ and represents $\bar{\varphi}$

As this diagram commutes we see:

• $\alpha=\theta\circ i$

and such, there's hopefully a nice way of showing that $\theta$ is an isomorphism.

Second Isomorphism Theorem

Let $(A,*)$ be a group, suppose that $N$ and $N'$ are normal subgroups of $A$ and that $N'\subseteq N$ then $N'$ is a normal subgroup of $N$, $N/N'$ is a normal subgroup of $A/N$ and:

• $$A/N\cong\frac{A/N'}{N/N'}$$

In fact there is an isomorphism, $\theta:A/N\rightarrow (A/N')/(N/N')$ such that:

• $\theta\circ\rho=\tau\circ\pi$ where $\pi$, $\rho$ and $\tau$ are canonical projections
  • Take this diagram: $\xymatrix{ A \ar[r]^\pi \ar[d]_{\rho} & A/N' \ar[d]^{\tau} \ar@{-->}[dl]^\sigma \\ A/N \ar@{.>}[r]^-\theta & (A/N')/(N/N') }$
    • if we first factor $\rho$ through $\pi$ to get $\sigma$

Third Isomorphism Theorem

Let $A$ be a subgroup of a group, $(G,*)$, let $N$ be a normal subgroup of $G$. Then:

1. $AN$ is a subgroup of $G$
2. $N$ is a normal subgroup of $AN$
3. $A\cap N$ is a normal subgroup of $A$ and
4. $AN/N\cong A/(A\cap N)$

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TODO: Finish off

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