Group Homomorphism, Conjugate, Center, and Abelian group
Problem 209
Let $G$ be a group. We fix an element $x$ of $G$ and define a map
\[ \Psi_x: G\to G\]
by mapping $g\in G$ to $xgx^{-1} \in G$.
Then prove the followings. (a) The map $\Psi_x$ is a group homomorphism.
(b) The map $\Psi_x=\id$ if and only if $x\in Z(G)$, where $Z(G)$ is the center of the group $G$.
(c) The map $\Psi_y=\id$ for all $y\in G$ if and only if $G$ is an abelian group.
For any elements $g, h\in G$, we have
\begin{align*}
\Psi_x(gh)=x(gh)x^{-1}\stackrel{(*)}{=} xgx^{-1}xhx^{-1}=\Psi_x(g) \Psi_x(h),
\end{align*}
where we inserted the identity element $e=x^{-1}x$ between $g$ and $h$ to obtain (*).
Hence $\Psi_x$ is a group homomorphism.
(b) The map $\Psi_x=\id$ if and only if $x\in Z(G)$
$(\implies)$ Suppose that $\Psi_x=\id$. Then for any $g\in G$, we have
\begin{align*}
\Psi_x(g)=\id(g)
\end{align*}
and thus we have
\begin{align*}
xgx^{-1}=g.
\end{align*}
This implies that we have $xg=gx$ for all $g \in G$, and hence $x\in Z(G)$.
$(\impliedby)$ On the other hand, if $x$ is in the center $Z(G)$, then we have
\[\Psi_x(g)=xgx^{-1}=xx^{-1}g=g\]
for any $g\in G$, where the second equality follows since $x \in Z(G)$.
This yields that $\Psi_x=\id$.
(c) The map $\Psi_y=\id$ for all $y\in G$ if and only if $G$ is an abelian group
$(\implies)$ Suppose that the map $\Psi_y=\id$ for all $y\in G$. Then by part (b), we have $y\in Z(G)$ for all $y\in G$. This means that we have $G=Z(G)$, and hence $G$ is an abelian group.
$(\impliedby)$ Now suppose that $G$ is an abelian group. Then for any $y\in G$ we have
\[\Psi_y(g)=ygy^{-1}=yy^{-1}g=g=\id(g)\]
for any $g\in G$, where the second equality follows since $G$ is an abelian group.
Thus we have $\Psi_y=\id$ for any $y \in G$.
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 […]
Cyclic Group if and only if There Exists a Surjective Group Homomorphism From $\Z$
Show that a group $G$ is cyclic if and only if there exists a surjective group homomorphism from the additive group $\Z$ of integers to the group $G$.
Proof.
$(\implies)$: If $G$ is cyclic, then there exists a surjective homomorhpism from $\Z$
Suppose that $G$ is […]
A Group Homomorphism and an Abelian Group
Let $G$ be a group. Define a map $f:G \to G$ by sending each element $g \in G$ to its inverse $g^{-1} \in G$.
Show that $G$ is an abelian group if and only if the map $f: G\to G$ is a group homomorphism.
Proof.
$(\implies)$ If $G$ is an abelian group, then $f$ […]
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 $h\in H$ […]
Group Homomorphism Sends the Inverse Element to the Inverse Element
Let $G, G'$ be groups. Let $\phi:G\to G'$ be a group homomorphism.
Then prove that for any element $g\in G$, we have
\[\phi(g^{-1})=\phi(g)^{-1}.\]
Definition (Group homomorphism).
A map $\phi:G\to G'$ is called a group homomorphism […]
A Group Homomorphism is Injective if and only if the Kernel is Trivial
Let $G$ and $H$ be groups and let $f:G \to K$ be a group homomorphism. Prove that the homomorphism $f$ is injective if and only if the kernel is trivial, that is, $\ker(f)=\{e\}$, where $e$ is the identity element of $G$.
Definitions/Hint.
We recall several […]