Tagged: group homomorphism

Group of $p$-Power Roots of 1 is Isomorphic to a Proper Quotient of Itself

Problem 221

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 $G$ itself.

 
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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.

 
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Isomorphism Criterion of Semidirect Product of Groups

Problem 113

Let $A$, $B$ be groups. Let $\phi:B \to \Aut(A)$ be a group homomorphism.
The semidirect product $A \rtimes_{\phi} B$ with respect to $\phi$ is a group whose underlying set is $A \times B$ with group operation
\[(a_1, b_1)\cdot (a_2, b_2)=(a_1\phi(b_1)(a_2), b_1b_2),\] where $a_i \in A, b_i \in B$ for $i=1, 2$.

Let $f: A \to A’$ and $g:B \to B’$ be group isomorphisms. Define $\phi’: B’\to \Aut(A’)$ by sending $b’ \in B’$ to $f\circ \phi(g^{-1}(b’))\circ f^{-1}$.

\[\require{AMScd}
\begin{CD}
B @>{\phi}>> \Aut(A)\\
@A{g^{-1}}AA @VV{\sigma_f}V \\
B’ @>{\phi’}>> \Aut(A’)
\end{CD}\] Here $\sigma_f:\Aut(A) \to \Aut(A’)$ is defined by $ \alpha \in \Aut(A) \mapsto f\alpha f^{-1}\in \Aut(A’)$.
Then show that
\[A \rtimes_{\phi} B \cong A’ \rtimes_{\phi’} B’.\]

 
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Basic Properties of Characteristic Groups

Problem 22

Definition (automorphism).

An isomorphism from a group $G$ to itself is called an automorphism of $G$.
The set of all automorphism is denoted by $\Aut(G)$.

Definition (characteristic subgroup).

A subgroup $H$ of a group $G$ is called characteristic in $G$ if for any $\phi \in \Aut(G)$, we have $\phi(H)=H$. In words, this means that each automorphism of $G$ maps $H$ to itself.

Prove the followings.

(a) If $H$ is characteristic in $G$, then $H$ is a normal subgroup of $G$.

(b) If $H$ is the unique subgroup of $G$ of a given order, then $H$ is characteristic in $G$.

(c) Suppose that a subgroup $K$ is characteristic in a group $H$ and $H$ is a normal subgroup of $G$. Then $K$ is a normal subgroup in $G$.

 
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