First Isomorphism Theorem/Groups
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Theorem
Let $\phi: G_1 \to G_2$ be a group homomorphism.
Let $\ker \left({\phi}\right)$ be the kernel of $\phi$.
Then:
- $\operatorname {Im} \left({\phi}\right) \cong G_1 / \ker \left({\phi}\right)$
where $\cong$ denotes group isomorphism.
Proof
Let $K = \ker \left({\phi}\right)$.
By Kernel is Normal Subgroup of Domain, $G_1 / K$ exists.
We need to establish that the mapping $\theta: G_1 / K \to G_2$ defined as:
- $\forall x \in G_1: \theta \left({x K}\right) = \phi \left({x}\right)$
is well-defined.
That is, we need to ensure that:
- $\forall x, y \in G: x K = y K \implies \theta \left({x K}\right) = \theta \left({y K}\right)$
Let $x, y \in G: x K = y K$. Then:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle x K\) | \(=\) | \(\displaystyle y K\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle x^{-1} y\) | \(\in\) | \(\displaystyle K = \ker \left({\phi}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Equal Cosets iff Product with Inverse in Coset‎ | ||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle \phi \left({x^{-1} y}\right)\) | \(=\) | \(\displaystyle e_{G_2}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle \left({\phi \left({x}\right)}\right)^{-1} \phi \left({y}\right)\) | \(=\) | \(\displaystyle e_{G_2}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle \phi \left({x}\right)\) | \(\in\) | \(\displaystyle \phi \left({y}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
Thus we see that $\theta$ is well-defined.
Since we also have that $\phi \left({x}\right) = \phi \left({y}\right) \implies x K = y K$, it follows that $\theta \left({x K}\right) = \theta \left({y K}\right) \implies x K = y K$.
So $\theta$ is injective.
We also note that $\operatorname {Im} \left({\theta}\right) = \left\{{\theta \left({x K}\right): x \in G}\right\}$.
So:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \operatorname {Im} \left({\theta}\right)\) | \(=\) | \(\displaystyle \left\{ {\theta \left({x K}\right): x \in G}\right\}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left\{ {\phi \left({x}\right): x \in G}\right\}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \operatorname {Im} \left({\phi}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
We also note that $\theta$ is a homomorphism:
| \(\displaystyle \) | \(\displaystyle \forall x, y \in G:\) | \(\displaystyle \) | \(\displaystyle \) | \(\) | \(\displaystyle \theta \left({x K y K}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \theta \left({x y K}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \phi \left({x y}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \phi \left({x}\right) \phi \left({y}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \theta \left({x K}\right) \theta \left({y K}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
Thus $\theta$ is a monomorphism whose image equals $\operatorname {Im} \left({\phi}\right)$.
The result follows.
$\blacksquare$
Also known as
Some sources call this the homomorphism theorem.
Others combine this result with Group Homomorphism Preserves Subgroups, Kernel is Subgroup and Kernel is Normal Subgroup of Domain.
Still others do not assign a special name to this theorem at all.
Sources
- J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 7.4$
- Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 66$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 52.1$
- John F. Humphreys: A Course in Group Theory (1996): $\S 8$: Theorem $8.13$