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.

Let $\R^{\times}=\R\setminus \{0\}$ be the multiplicative group of real numbers.
Let $\C^{\times}=\C\setminus \{0\}$ be the multiplicative group of complex numbers.
Then show that $\R^{\times}$ and $\C^{\times}$ are not isomorphic as groups.

Recall.

Let $G$ and $K$ be groups.
Recall that a map $f: G \to K$ is a group homomorphism if
\[f(ab)=f(a)f(b)\] for all $a, b \in G$.
A group isomorphism is a bijective homomorphism.
If there is a group isomorphism from $G$ to $K$, we say that $G$ and $K$ are isomorphic (as groups).
 
We give two proofs.

Proof 1.

Seeking a contradiction, assume that there is a group isomorphism $\phi: \C^{\times} \to \R^{\times}$.
Since $\phi$ is a group homomorphism, $\phi(1)=1$. Thus we have
\[1=\phi(1)=\phi((-1)(-1))=\phi(-1)\phi(-1)=\phi(-1)^2.\] Hence $\phi(-1)=\pm 1$. But since $\phi$ is injective and $\phi(1)=1$, we must have $\phi(-1)=-1$.

Now we have
\begin{align*}
-1=\phi(-1)=\phi(i^2)=\phi(i)^2.
\end{align*}

Since $\phi(i)\in \R^{\times}$, $\phi(i)^2$ must be a positive number. Thus we reached a contradiction.
Hence there is no isomorphism between $\R^{\times}$ and $\C^{\times}$.

Proof 2.

Suppose that there is a group isomorphism $\phi: \C^{\times} \to \R^{\times}$.
We want to find a contradiction.
Let $\zeta=e^{2\pi i /3}$ be a primitive third root of unity.

Since $\zeta^3=1$ and $\phi$ is a group homomorphism, we have
\begin{align*}
1=\phi(1)=\phi(\zeta^3)=\phi(\zeta)^3.
\end{align*}
Since $\phi(\zeta)$ is a real number, this implies that $\phi(\zeta)=1$.

This is a contradiction since $\phi$ is injective, but we have $\phi(1)=1=\phi(\zeta)$.
Therefore, there cannot be a group isomorphism between $\R^{\times}$ and $\C^{\times}$.

The post Multiplicative Groups of Real Numbers and Complex Numbers are not Isomorphic first appeared on Problems in Mathematics.