Let $R$ be a commutative ring with $1$ and let $M$ be an $R$-module.
Prove that the $R$-module $M$ is irreducible if and only if $M$ is isomorphic to $R/I$, where $I$ is a maximal ideal of $R$, as an $R$-module.

Definition (Irreducible module).

An $R$-module $M$ is called irreducible if $M$ is not the zero module and $0$ and $M$ are the only submodules of $M$.

An irreducible module
is also called a simple module.

Proof.

$(\implies)$ Suppose that $M$ is an irreducible $R$-module.
Then by definition of an irreducible module, $M$ is not the zero module.
Take any nonzero element $m\in M$, and consider the cyclic submodule $(m)=Rm$ generated by $m$.
Since $M$ is irreducible, we must have $M=Rm$.

Now we define a map $f:R\to M$ by sending $r\in R$ to $f(r)=rm$.
Then the map $f$ is an $R$-module homomorphism regarding $R$ is an $R$-module.

In fact, we have
\begin{align*}
&f(r+s)=(r+s)m=rm+sm=f(r)+f(s)\\
&f(rs)=(rs)m=r(sm)=rf(s)
\end{align*}
for any $r, s\in R$.
Since $M=Rm$, the homomorphism $f$ is surjective.
Thus, by the first isomorphism theorem, we obtain
\[R/I\cong M,\] where $I=\ker(f)$.

It remains to show that $I$ is a maximal ideal of $R$.
Suppose that $J$ is an ideal such that $I\subset J \subset R$.
Then by the third isomorphism theorem for rings, we know that $J/I$ is an ideal of the ring $R/I\cong M$, hence $J/I$ is a submodule.

Since $M$ is irreducible, we must have either $J/I=0$ or $J/I=M$.
This implies that $J=I$ or $J=M$.
Hence $I$ is a maximal ideal.

$(\impliedby)$ Suppose now that $M\cong R/I$ for some maximal ideal $I$ of $R$.
Let $N$ be any submodule of $R/I$. (We identified $M$ and $R/I$ by the above isomorphism.)
Then $N$ is an ideal of $R/I$ since $N$ is an abelian group and closed under the action of $R$, hence that of $R/I$.

Since $R$ is a commutative ring and $I$ is a maximal ideal of $R$, we know that $R/I$ is a field.
Thus, only the ideals of $R/I$ are $0$ or $R/I$.

Hence we have $N=0$ or $N=R/I=M$.
This proves that $M$ is irreducible.

Related Question.

A similar technique in the proof above can be used to solve the following problem.

See the post “A module is irreducible if and only if it is a cyclic module with any nonzero element as generator” for a proof.

The post A Module $M$ is Irreducible if and only if $M$ is isomorphic to $R/I$ for a Maximal Ideal $I$. first appeared on Problems in Mathematics.

Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$.
Let $\Aut(N)$ be the group of automorphisms of $G$.

Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime.
Then prove that $N$ is contained in the center of $G$.

Outline of the proof

Here is the outline of the proof.

1. Define a group homomorphism $\psi: G\to \Aut(N)$ by $\psi(g)(n)=gng^{-1}$ for all $g\in G$ and $n\in N$.
   We need to check:
• The map $\psi(g)$ is an automorphism of $N$ for each $g\in G$.
• The map $\psi$ is in fact a group homomorphism from $G$ to $\Aut(N)$.
2. The assumption that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime implies that $G=\ker(\psi)$.
3. This implies that $N$ is in the center of $G$.

Proof.

We define a group homomorphism $\psi: G \to \Aut(N)$ as follows.
For each $g\in G$, we first define an automorphism $\psi(g)$ of $N$.
Define $\psi(g): N \to N$ by
\[\psi(g)(n)=gng^{-1}.\]

Note that since $N$ is a normal subgroup of $G$, the output $\psi(g)(n)=gng^{-1}$ actually lies in $N$.

We prove that so defined $\psi(g)$ is a group homomorphism from $N$ to $N$ for each fixed $g\in G$.
For $n_1, n_2 \in N$, we have
\begin{align*}
\psi(g)(n_1n_2)&=g(n_1n_2)g^{-1} && \text{by definition of $\psi(g)$}\\
&=gn_1g^{-1}gn_2g^{-1} && \text{by inserting $e=g^{-1}g$}\\
&=\psi(g)(n_1) \psi(g)(n_2) && \text{by definition of $\psi(g)$}.
\end{align*}
It follows that $\psi(g)$ is a group homomorphism, and hence $\psi(g)\in \Aut(N)$.

We have defined a map $\psi:G\to \Aut(N)$. We now prove that $\psi$ is a group homomorphism.
For any $g_1, g_2$, and $n\in N$, we have
\begin{align*}
\psi(g_1 g_2)(n)&=(g_1g_2)n(g_1 g_2)^{-1}\\
&=g_1 g_2 n g_2^{-1} g_1^{-1}\\
&=g_1 \psi(g_2)(n) g_1^{-1}\\
&=\psi(g_1)\psi(g_2)(n).
\end{align*}

Thus, $\psi: G\to \Aut(N)$ is a group homomorphism.
By the first isomorphism theorem, we have
\[G/\ker(\psi)\cong \im(\psi)< \Aut(N). \tag{*}\] Note that if $g\in N$, then $\psi(g)(n)=gng^{-1}=n$ since $N$ is abelian. It yields that the subgroup $N$ is in the kernel $\ker(\psi)$.

Then by the third isomorphism theorem, we have
\begin{align*}
G/\ker(\psi) \cong (G/N)/(\ker(\psi)/N). \tag{**}
\end{align*}

It follows from (*) and (**) that the order of $G/\ker(\psi)$ divides both the order of $\Aut(N)$ and the order of $G/N$. Since the orders of the latter two groups are relatively prime by assumption, the order of $G/\ker(\psi)$ must be $1$. Thus the quotient group is trivial and we have
\[G=\ker(\psi).\]

This means that for any $g\in G$, the automorphism $\psi(g)$ is the identity automorphism of $N$.
Thus, for any $g\in G$ and $n\in N$, we have $\psi(g)(n)=n$, and thus $gng^{-1}=n$.
As a result, the subgroup $N$ is contained in the center of $G$.

The post Abelian Normal subgroup, Quotient Group, and Automorphism Group first appeared on Problems in Mathematics.

Let $H$ and $K$ be normal subgroups of a group $G$.
Suppose that $H < K$ and the quotient group $G/H$ is abelian.
Then prove that $G/K$ is also an abelian group.

Solution.

We will give two proofs.

Hint (The third isomorphism theorem)

Recall the third isomorphism theorem of groups:
Let $G$ be a group and let $H, K$ be normal subgroups of $G$ with $H < K$.
Then we have $G/K$ is a normal subgroup of $G/H$ and we have an isomorphism
\[G/K \cong (G/H)/(G/K).\]

Proof 1 (Using third isomorphism theorem)

Since $H, K$ are normal subgroups of $G$ and $H < K$, the third isomorphism theorem yields that
\[G/K \cong (G/H)/(G/K).\]

Since the group $G/H$ is abelian by assumption, and in general a quotient group of an abelian group is abelian, it follows $(G/H)/(G/K)$ is an abelian group.

Hence by the above isomorphism, the group $G/K$ is also an abelian group.

Proof 2 (Using the commutator subgroup)

Here is another proof using the commutator subgroup $[G, G]$ of $G$.
Recall that for a subgroup $N$ of $G$, the following two conditions are equivalent.

1. The subgroup $N$ is normal and the $G/N$ is an abelian.
2. The commutator subgroup $[G, G]$ is a subgroup of $N$.

For the proof of this fact, see the post “Commutator subgroup and abelian quotient group“.

Now we prove the problem using this fact.
Since $H$ is normal and the quotient $G/H$ is an abelian group, the commutator subgroup $[G, G]$ is a subgroup of $H$ by the fact (1 $\implies$ 2).

Then we have
\begin{align*}
[G, G] < H < K.
\end{align*}
Hence $[G, G]$ is a subgroup of $K$, hence $G/K$ is an abelian group by the fact again (2 $\implies$ 1).

The post If Quotient $G/H$ is Abelian Group and $H < K \triangleleft　G$, then $G/K$ is Abelian first appeared on Problems in Mathematics.