Some Applications of the First Isomorphism Theorem and the Correspondence Theorem

Proposition 1. Let $N$ be a normal subgroup of a group $G$. Then $N$ is a maximal normal subgroup of $G$ if and only if $G/N$ is simple.

Proof : Consider the natural homomorphism
$$
\pi:G\longrightarrow G/N
$$
given by
$$
g\longmapsto gN.
$$
Its kernel is
$$
\operatorname{Ker}\pi=\{g\in G\mid gN=N\}=N.
$$

By the Correspondence Theorem, the normal subgroups of $G/N$ correspond exactly to the normal subgroups of $G$ containing $N$. Therefore, $N$ is a maximal normal subgroup of $G$.

if and only if the only normal subgroups of $G$ containing $N$ are $N$ and $G$, which holds if and only if the only normal subgroups of $G/N$ are
$$
N/N \quad \text{and} \quad G/N,
$$
that is, if and only if $G/N$ is simple.

Proposition 2. Let $N$ and $H$ be two distinct maximal normal subgroups of a group $G$. Then
$$
H\cap N
$$
is a maximal normal subgroup of $H$.

Proof : By Proposition 1, it suffices to prove that
$$
H/(H\cap N)
$$
is simple.

By the Second Isomorphism Theorem,
$$
H/(H\cap N)\cong HN/N.
$$

We claim that
$$
HN\neq N.
$$
Indeed, if $HN=N$, then for every $h\in H$ we would have
$$
hN\subseteq HN=N.
$$
Since also $N\subseteq hN$, it follows that
$$
hN=N,
$$
so
$$
h\in N.
$$
Hence
$$
H\subseteq N.
$$
Because $H\neq N$ and $H$ is a maximal normal subgroup of $G$, this would force
$$
N=G,
$$
contradicting the fact that $N$ is a maximal normal subgroup of $G$.

Thus
$$
HN\neq N.
$$
Since both $H$ and $N$ are normal subgroups of $G$, the product $HN$ is also a normal subgroup of $G$. Moreover,
$$
N\subsetneqq HN.
$$
Because $N$ is a maximal normal subgroup of $G$, we must have
$$
HN=G.
$$
Therefore,
$$
H/(H\cap N)\cong G/N.
$$
Since $N$ is a maximal normal subgroup of $G$, Proposition 1 implies that $G/N$ is simple. Hence $H/(H\cap N)$ is simple. Applying Proposition 1 again inside the group $H$, we conclude that
$$
H\cap N
$$
is a maximal normal subgroup of $H$.

Proposition 3. Let $G$ and $H$ be two cyclic groups with
$$
|G|=m,\qquad |H|=n.
$$
Then there exists a surjective homomorphism from $G$ onto $H$ if and only if
$$
n\mid m.
$$

Proof : First assume that there exists a surjective homomorphism
$$
\varphi:G\longrightarrow H.
$$
By the First Isomorphism Theorem,
$$
G/\operatorname{Ker}\varphi\cong H.
$$
Hence
$$
|H|=|G/\operatorname{Ker}\varphi|=\frac{|G|}{|\operatorname{Ker}\varphi|}.
$$
Therefore $n$ divides $m$, that is,
$$
n\mid m.
$$

Conversely, assume that
$$
n\mid m.
$$
Since $G$ is a cyclic group of order $m$ and $H$ is a cyclic group of order $n$, there exist isomorphisms
$$
\sigma:G\longrightarrow \mathbb{Z}_m,\qquad \tau:\mathbb{Z}_n\longrightarrow H.
$$
Define a map
$$
\varphi:\mathbb{Z}_m\longrightarrow \mathbb{Z}_n
$$
by
$$
\bar a\longmapsto \tilde a.
$$
We verify that $\varphi$ is a well-defined surjective homomorphism.

To see that it is well-defined, suppose that
$$
\bar a=\bar b
$$
in $\mathbb{Z}_m$. Then
$$
m\mid (b-a).
$$
Since $n\mid m$, it follows that
$$
n\mid (b-a),
$$
and hence
$$
\tilde a=\tilde b
$$
in $\mathbb{Z}_n$.

Next, for any $\bar a,\bar b\in \mathbb{Z}_m$, we have
$$
\varphi(\bar a+\bar b)=\varphi(\overline{a+b})=\widetilde{a+b}=\tilde a+\tilde b=\varphi(\bar a)+\varphi(\bar b),
$$
so $\varphi$ is a homomorphism. It is clearly surjective.

Therefore the composition
$$
f=\tau\circ \varphi\circ \sigma:G\longrightarrow H
$$
is a surjective homomorphism.

Proposition 4. Every homomorphism from a finite cyclic group to an infinite cyclic group is necessarily trivial.

Proof : Let $H$ be a finite cyclic group, let $G$ be an infinite cyclic group, and let
$$
\varphi:H\longrightarrow G
$$
be a homomorphism. Consider the image
$$
\operatorname{Im}\varphi.
$$
Since $\operatorname{Im}\varphi$ is a subgroup of $G$, and $G$ is an infinite cyclic group, it follows that if
$$
\operatorname{Im}\varphi\neq \{e\},
$$
then $\operatorname{Im}\varphi$ is itself an infinite cyclic group.

By the First Isomorphism Theorem,
$$
H/\operatorname{Ker}\varphi\cong \operatorname{Im}\varphi.
$$
Thus $H/\operatorname{Ker}\varphi$ would be an infinite cyclic group. However, since $H$ is finite, we have
$$
|H/\operatorname{Ker}\varphi|=\frac{|H|}{|\operatorname{Ker}\varphi|}<\infty,
$$
which is impossible.

Therefore,
$$
\operatorname{Im}\varphi=\{e\}.
$$
Hence $\varphi$ is the trivial homomorphism.

The cover image of this article was taken at Lake Baikal in Russia.