Abelian Normal subgroup, Quotient Group, and Automorphism Group

Abelian Group problems and solutions

Problem 343

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 $G$.

 
LoadingAdd to solve later
Sponsored Links
 

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 $G$.


LoadingAdd to solve later

Sponsored Links

More from my site

  • Group of $p$-Power Roots of 1 is Isomorphic to a Proper Quotient of ItselfGroup 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 $G$ […]
  • Subgroup of Finite Index Contains a Normal Subgroup of Finite IndexSubgroup 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 Homomorphisms From Group of Order 21 to Group of Order 49Group Homomorphisms From Group of Order 21 to Group of Order 49 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 $K$.   Proof. Let $e$ be the identity element of the group […]
  • Group Homomorphism, Conjugate, Center, and Abelian groupGroup Homomorphism, Conjugate, Center, and Abelian group 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 show that (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 […]
  • Equivalent Definitions of Characteristic Subgroups. Center is Characteristic.Equivalent Definitions of Characteristic Subgroups. Center is Characteristic. Let $H$ be a subgroup of a group $G$. We call $H$ characteristic in $G$ if for any automorphism $\sigma\in \Aut(G)$ of $G$, we have $\sigma(H)=H$. (a) Prove that if $\sigma(H) \subset H$ for all $\sigma \in \Aut(G)$, then $H$ is characteristic in $G$. (b) Prove that the center […]
  • Surjective Group Homomorphism to $\Z$ and Direct Product of Abelian GroupsSurjective Group Homomorphism to $\Z$ and Direct Product of Abelian Groups Let $G$ be an abelian group and let $f: G\to \Z$ be a surjective group homomorphism. Prove that we have an isomorphism of groups: \[G \cong \ker(f)\times \Z.\]   Proof. Since $f:G\to \Z$ is surjective, there exists an element $a\in G$ such […]
  • Basic Properties of Characteristic GroupsBasic Properties of Characteristic Groups 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 […]
  • Group Homomorphism, Preimage, and Product of GroupsGroup 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 $h\in H$ […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Group Theory
Abelian Group problems and solutions
Surjective Group Homomorphism to $\Z$ and Direct Product of Abelian Groups

Let $G$ be an abelian group and let $f: G\to \Z$ be a surjective group homomorphism. Prove that we have...

Close