Let $f: G \rightarrow H$ be a homomorphism. Then we have a normal subgroup
$$
\operatorname{Ker}(f) \subset G
$$
thus we can also define the quotient group
$$
G / \operatorname{Ker}(f)
$$
One might ask
Is there a relation between the groups $G / \operatorname{Ker}(f)$ and $H$ and homomorphism $H ?$
The answer to this question is given by the following fundamental result.
3. Theorem 1: First Isomorphism Theorem
Let $f: G \rightarrow H$ be a homomorphism. Let
$$
\pi: G \rightarrow G / \operatorname{Ker}(f)
$$
be the natural surjective homomorphism sending every element $x \in G$ to the coset containing $x .$ Then there exists a unique isomorphism
$$
\widetilde{f}: G / \operatorname{Ker}(f) \rightarrow \operatorname{Im}(f) \subset H
$$
such that for any $x \in G$
$$
f(x)=(\widetilde{f} \circ \pi)(x) .
$$
Proof. From the previous lecture we know that cosets of the kernel Ker $(f) \subset G$ are in 1 -to- 1 correspondence with the elements in $\operatorname{Im}(f)$. Thus we have a well-defined bijection
$$
\widetilde{f}: G / \operatorname{Ker}(f) \rightarrow \operatorname{Im}(f)
$$
which sends a coset $A \subset G$ to $f(a) \in \operatorname{Im}(a)$, where $a$ is any representative of $A .$
We need to check that $\widetilde{f}$ is a homomorphism. Indeed, if $x \operatorname{Ker}(f)$ and $y \operatorname{Ker}(f)$ are any two elements of $G / \operatorname{Ker}(f)$, then
$$
\widetilde{f}(x \operatorname{Ker}(f) \cdot y \operatorname{Ker}(f))=^{1} \widetilde{f}(x y \operatorname{Ker} f)=^{2} f(x y)=^{3} f(x) f(y)=^{4} \widetilde{f}(x \operatorname{Ker}(f)) \widetilde{f}(y \operatorname{Ker}(f))
$$
where in the first identity we used the definition of the multiplication in the quotient group G/Ker $(f)$, in the second identity we used the definition of $\widetilde{f}$, in the identity 3 we used the fact that $f$ is a homomorphism, and in the last identity 4 we again used the definition of $\widetilde{f}$.
Alternatively, we can formulate the First Isomorphism Theorem via a commutative diagram:
which says that given a homomorphism $f: G \rightarrow H$ and the corresponding surjective homomorphism $\pi: G \rightarrow$ $G / \operatorname{Ker}(f)$, there exists a unique isomorphism $f: G / \operatorname{Ker}(f) \rightarrow \operatorname{Im}(f)$ such that the above diagram is commutative, i.e., $f=f \circ \pi$.
In many situations, the first isomorphism theorem allows us to get a better understanding of the quotient groups $G / N$. Specifically, if we want to describe $G / N$ explicitly, often it is helpful to find a homomorphism
$$
f: G \rightarrow H
$$
to some other group $H$, such that $\operatorname{Ker}(f)=N .$ Then, by the theorem $G / N$ is isomorphic to Im $(f) .$
- Consider $G=S_{n}$ and its normal subgroup $N=A_{n} .$ Then we can realize $A_{n}$ as the kernel of the sign homomorphism
$$
\text { sgn: } S_{n} \rightarrow{+1,-1}
$$
Thus
$$
A_{n} / S_{n}=\operatorname{Im}(\operatorname{sgn})={+1,-1}
$$
and the latter group is isomorphic to $\mathbb{Z}_{2}$.
- Let $G=G L_{2}(\mathbb{R})$ be the group of $2 \times 2$ matrices with nonzero determinant. Consider its subgroup
$$
S L_{2}(\mathbb{R})=\{A \in G L_{2}(\mathbb{R}) \mid \operatorname{det}(A)=1\} .
$$
This group is the kernel of the determinant homomorphism
$$
\operatorname{det}: G L_{2}(\mathbb{R}) \rightarrow \mathbb{R} \backslash \{0\}
$$
Thus, since det is surjective (i.e., $\operatorname{Im}($ det $)=\mathbb{R} \backslash \{0\} )$, we deduce from the isomorphism theorem
$$
G L_{2}(\mathbb{R}) / S L_{2}(\mathbb{R}) \simeq \mathbb{R} \backslash \{0\} .
$$