Let $G$ be a finite group of order $21$ and let $K$ be a finite group of order $49$.
Suppose that $G$ does not have a normal subgroup of order $3$.
Then determine all group homomorphisms from $G$ to $K$.

Let $e$ be the identity element of the group $K$.
We claim that every group homomorphism from $G$ to $K$ is trivial.
Namely, if $\phi:G \to K$ is a group homomorphism, then we have $\phi(g)=e$ for every $g\in G$.

The first isomorphism theorem gives the isomorphism
\[G/\ker(\phi)\cong \im(\phi) < K.\]
It follows that the order $|\im(\phi)|$ of the image $\im(\phi)$ is a divisor of the order of $G$ and that of $K$.
Hence the order $|\im(\phi)|$ divides the greatest common divisor of $|G|=3\cdot 7$ and $|K|=7^2$, which is $7$.
So, the possibilities are $|\im(\phi)|=1, 7$.
If $|\im(\phi)|=7$, then we have
\[\frac{|G|}{|\ker(\phi)|}=|\im(\phi)|=7,\]
and we obtain $|\ker(\phi)|=3$.
Since the kernel of a group homomorphism is a normal subgroup, this contradicts the assumption that $G$ does not have a normal subgroup of order $3$.
Therefore, we must have $|\im(\phi)|=1$, and this implies that $\phi$ is a trivial homomorphism.
Thus we conclude that every group homomorphism from $G$ to $K$ is trivial.

Abelian Normal subgroup, Quotient Group, and Automorphism Group
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 […]

Subgroup of Finite Index Contains a Normal Subgroup of Finite Index
Let $G$ be a group and let $H$ be a subgroup of finite index. Then show that there exists a normal subgroup $N$ of $G$ such that $N$ is of finite index in $G$ and $N\subset H$.
Proof.
The group $G$ acts on the set of left cosets $G/H$ by left multiplication.
Hence […]

Group Homomorphism, Preimage, and Product of Groups
Let $G, G'$ be groups and let $f:G \to G'$ be a group homomorphism.
Put $N=\ker(f)$. Then show that we have
\[f^{-1}(f(H))=HN.\]
Proof.
$(\subset)$ Take an arbitrary element $g\in f^{-1}(f(H))$. Then we have $f(g)\in f(H)$.
It follows that there exists $h\in H$ […]

The Set of Square Elements in the Multiplicative Group $(\Zmod{p})^*$
Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$.
(a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$.
(b) Determine the index $[G : S]$.
(c) Assume […]

Group Homomorphism from $\Z/n\Z$ to $\Z/m\Z$ When $m$ Divides $n$
Let $m$ and $n$ be positive integers such that $m \mid n$.
(a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined.
(b) Prove that $\phi$ is a group homomorphism.
(c) Prove that $\phi$ is surjective.
(d) Determine […]

Every Group of Order 24 Has a Normal Subgroup of Order 4 or 8
Prove that every group of order $24$ has a normal subgroup of order $4$ or $8$.
Proof.
Let $G$ be a group of order $24$.
Note that $24=2^3\cdot 3$.
Let $P$ be a Sylow $2$-subgroup of $G$. Then $|P|=8$.
Consider the action of the group $G$ on […]

Group of $p$-Power Roots of 1 is Isomorphic to a Proper Quotient of Itself
Let $p$ be a prime number. Let
\[G=\{z\in \C \mid z^{p^n}=1\} \]
be the group of $p$-power roots of $1$ in $\C$.
Show that the map $\Psi:G\to G$ mapping $z$ to $z^p$ is a surjective homomorphism.
Also deduce from this that $G$ is isomorphic to a proper quotient of $G$ […]

A Group Homomorphism is Injective if and only if Monic
Let $f:G\to G'$ be a group homomorphism. We say that $f$ is monic whenever we have $fg_1=fg_2$, where $g_1:K\to G$ and $g_2:K \to G$ are group homomorphisms for some group $K$, we have $g_1=g_2$.
Then prove that a group homomorphism $f: G \to G'$ is injective if and only if it is […]