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 $G$.

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 $G$.

The post Abelian Normal subgroup, Quotient Group, and Automorphism Group first appeared on Problems in Mathematics.

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 $Z(G)$ of $G$ is characteristic in $G$.

Definition

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

Proof.

(a) If $\sigma(H) \subset H$ for all $\sigma \in \Aut(G)$, then $H$ is characteristic in $G$

Since $\sigma$ is an automorphism, the inverse $\sigma^{-1}$ is also an automorphism of $G$.
Hence, we have
\[\sigma^{-1}(H)\subset H\] by the assumption.

Applying $\sigma$, we have
\[\sigma\sigma^{-1}(H) \subset \sigma(H).\] Then we obtain
\begin{align*}
H&=\sigma \sigma^{-1}(H)\subset \sigma(H)\subset H.
\end{align*}

Since the both ends are $H$, the inclusion is in fact the equality.
Thus, we obtain
\[\sigma(H)=H,\] and the subgroup $H$ is characteristic in the group $G$.

(b) The center $Z(G)$ of $G$ is characteristic in $G$

By part (a), it suffices to prove that $\sigma(Z(G)) \subset Z(G)$ for every automorphism $\sigma \in \Aut(G)$ of $G$.

Let $x\in \sigma(Z(G))$. Then there exists $y \in Z(G)$ such that $x=\sigma(y)$.
To show that $x \in Z(G)$, consider an arbitrary $g \in G$.
Then since $\sigma$ is an automorphism, we have $G=\sigma(G)$.
Thus there exists $g’$ such that $g=\sigma(g’)$.

We have
\begin{align*}
xg &=\sigma(y)\sigma(g’)\\
&=\sigma(yg’) && \text{ (since $\sigma$ is a homomorphism)}\\
&=\sigma(g’y) && \text{ (since $y \in Z(G)$)}\\
&=\sigma(g’)\sigma(y) && \text{ (since $\sigma$ is a homomorphism)}\\
&=gx.
\end{align*}
Since this is true for all $g \in G$, it follows that $x \in Z(G)$, and thus
\[\sigma(Z(G)) \subset Z(G).\] This completes the proof.

Comment.

In some textbook, a subgroup $H$ of $G$ is said to be characteristic in $G$ if $\sigma(H) \subset H$ for all $\sigma \in \Aut(G)$.
Problem (a) implies that our definition of characteristic and this alternative definition are in fact equivalent.

Related Question.

Read the post Basic properties of characteristic groups for more problems about characteristic subgroups.

The post Equivalent Definitions of Characteristic Subgroups. Center is Characteristic. first appeared on Problems in Mathematics.

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.

Proof.

(a) The map $\Psi_x$ is a group homomorphism

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 $y \in G$.

The post Group Homomorphism, Conjugate, Center, and Abelian group first appeared on Problems in Mathematics.

Let $p$ be a prime number.
Let $G$ be a non-abelian $p$-group.
Show that the index of the center of $G$ is divisible by $p^2$.

Proof.

Suppose the order of the group $G$ is $p^a$, for some $a \in \Z$.
Let $Z(G)$ be the center of $G$. Since $Z(G)$ is a subgroup of $G$, the order of the center is also a power of $p$, that is, $|Z(G)|=p^b$, for some $b \in \Z$.
Then we have the index $[G: Z(G)]=p^{a-b}$.

If $a-b=0$, then we have $G=Z(G)$ and $G$ is an abelian group. This contradicts with the assumption that $G$ is non-abelian. So $a-b \neq 0$.

If $a-b=1$, then the order of the quotient $|G/Z(G)|=[G:Z(G)]=p$ is a prime, thus $G/Z(G)$ is a cyclic group.
Recall that if the quotient by the center is cyclic, then the group is abelian.
Thus the group $G$ is abelian, which again a contradiction.

Therefore, we must have $a-b \geq 2$, hence $p^2$ divides the index $[G: Z(G)]=p^{a-b}$.
This concludes the proof.

The post The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$ first appeared on Problems in Mathematics.

Let $H$ be a subgroup of order $2$. Let $N_G(H)$ be the normalizer of $H$ in $G$ and $C_G(H)$ be the centralizer of $H$ in $G$.

(a) Show that $N_G(H)=C_G(H)$.

(b) If $H$ is a normal subgroup of $G$, then show that $H$ is a subgroup of the center $Z(G)$ of $G$.

Definitions.

Recall that the centralizer of $H$ in $G$ is
\[C_G(H)=\{g \in G \mid gh=hg \text{ for any } h\in H\}.\]

The normalizer of $H$ in $G$ is
\[N_G(H)=\{g\in G \mid gH=Hg \}.\]

Proof.

(a) Prove $N_G(H)=C_G(H)$

In general, we have $C_G(H) \subset N_G(H)$. We show that $N_G(H) \subset C_G(H)$.

Take any $g \in N_G(H)$. We have $gH=Hg$. Since $|H|=2$, let $H=\{1,h\}$.
Then $gH=\{g,gh\}$ and $Hg=\{g, hg\}$. Since $gH=Hg$, we have $gh=hg$. Namely $g\in C_G(H)$.

This proves that $N_G(H) \subset C_G(H)$, hence $N_G(H)= C_G(H)$.

(b) If $H$ is normal, then $H$ is a subgroup of $Z(G)$

Suppose that $H$ is a normal subgroup of $G$, that is $G=N_G(H)$.
By part (a), this implies that $G=C_G(H)$. Hence $H < Z(G)$.

The post Normalizer and Centralizer of a Subgroup of Order 2 first appeared on Problems in Mathematics.

Let $D_8$ be the dihedral group of order $8$.
Using the generators and relations, we have
\[D_{8}=\langle r,s \mid r^4=s^2=1, sr=r^{-1}s\rangle.\]

(a) Let $A$ be the subgroup of $D_8$ generated by $r$, that is, $A=\{1,r,r^2,r^3\}$.
Prove that the centralizer $C_{D_8}(A)=A$.

(b) Show that the normalizer $N_{D_8}(A)=D_8$.

(c) Show that the center $Z(D_8)=\langle r^2 \rangle=\{1,r^2\}$, the subgroup generated by $r^2$.

Definitions (centralizer, normalizer, center).

Recall the definitions.

1.  The centralizer $C_{D_8}(A)$ is a subgroup of $D_8$ whose elements commute with $A$.
   That is $C_{D_8}(A)=\{ g\in D_8 \mid gxg^{-1}=x \text{ for all } x\in A\}$.
2. The normalizer $N_{D_8}(A)$ is a subgroup of $D_8$ defined as
   $N_{D_8}(A)=\{ g\in D_8 \mid gxg^{-1} \in A \text{ for any } x \in A\}$.
3. The center $Z(D_8)$ is a subgroup of $D_8$ whose elements commute with all elements of $D_8$.
   That is, $Z(D_8)=\{g \in D_8 \mid gxg^{-1}=x \text{ for all } x\in D_8\}$.

Proof.

(a) The centralizer $C_{D_8}(A)=A$

 Since any power of $r$ commutes with each other we have $A < C_{D_8}(A)$.
Since $sr=r^{-1}s$ and $r^{-1}s\neq rs$ (otherwise $r^2=1$), we see that $s\not \in C_{D_8}(A)$.

This also implies that any element of the form $r^as \in D_8$ is not in $C_{D_8}(A)$. In fact, if $r^as\in C_{D_8}(A)$, then $s=(r^{-a})\cdot (r^a s)$ is also in $C_{D_8}(A)$ because $r^{-a}\in C_{D_8}(A)$ and $C_{D_8}(A)$ is a group.

This is a contradiction. Therefore $C_{D_8}(A)=A$.

(b) The normalizer $N_{D_8}(A)=D_8$

 In general, the centralizer of a subset is contained in the normalizer of the subset. From this fact we have $A=C_{D_8}(A) < N_{D_8}(A)$.
Thus it suffices to show that the other generator $s \in D_8$ belongs to $N_{D_8}(A)$.

We have $sr^as^{-1}=r^{-1}ss^{-1}=r^{-1}\in A$ using the relation $sr=r^{-1}s$.
Thus $s \in N_{D_8}(A)$ as required.

(c) The center $Z(D_8)=\langle r^2 \rangle=\{1,r^2\}$

The center $Z(D_8)$ is contained in the centralizer $C_{D_8}(A)=A$.
Since $rsr^{-1}=sr^2 \neq s$ and $r^3s(r^3)^{-1}=r^{-1}sr=sr^2\neq s$, the elements $r$ and $r^3$ are not in the center $Z(D_8)$.

On the other hand, we have $sr^2=r^{-1}sr=r^{-2}s=r^2s$. Thus $r^2s(r^2)^{-1}=s$ and $r^2 \in Z(D_8)$.
Therefore we have $Z(D_8)=\{1, r^2\}$.

The post Centralizer, Normalizer, and Center of the Dihedral Group $D_{8}$ first appeared on Problems in Mathematics.

Show that the center $Z(S_n)$ of the symmetric group with $n \geq 3$ is trivial.

Steps/Hint

1. Assume $Z(S_n)$ has a non-identity element $\sigma$.
2. Then there exist numbers $i$ and $j$, $i\neq j$, such that $\sigma(i)=j$
3. Since $n\geq 3$ there exists another number $k$.
4. Let $\tau=( i k)\in S_n$ and find a contradiction.

Proof.

Seeking a contradiction, assume that the center $Z(S_n)$ is non-trivial.
Then there exists a non-identity element $\sigma \in Z(G)$.

Since $\sigma$ is a non-identity element, there exist numbers $i$ and $j$, $i\neq j$, such that $\sigma(i)=j$.

Now by assumption $n \geq 3$, there exists another number $k$ that is different from $i$ and $j$. Let us consider the transposition $\tau=(i k)\in S_n$.
Then we have
\begin{align*}
\tau \sigma (i)&=\tau (j)=j \\
\sigma \tau (i) &= \sigma (k) \neq j
\end{align*}
since $\sigma(i)=j$ and $\sigma$ is bijective.

Thus $\tau \sigma \neq \sigma \tau$ but this contradicts that $\sigma \in Z(S_n)$.
Therefore $Z(S_n)$, $n \geq 3$, must be trivial.

The post The Center of the Symmetric group is Trivial if $n>2$ first appeared on Problems in Mathematics.

Let $G$ be a group of order $|G|=pq$, where $p$ and $q$ are (not necessarily distinct) prime numbers.

Then show that $G$ is either abelian group or the center $Z(G)=1$.

Hint.

Use the result of the problem “If the Quotient by the Center is Cyclic, then the Group is Abelian”.

Proof.

Since the center $Z(G)$ is a (normal) subgroup  of $G$, the order of $Z(G)$ divides the order of $G$ by Lagrange’s theorem.
Thus the order of $Z(G)$ is one of $1,p,q,pq$.

Suppose that $Z(G)\neq 1$.
Then the order of the quotient group $G/Z(G)$ is one of $1,p,q$.
Hence the group $G/Z(G)$ is a cyclic group.

We conclude that $G$ is abelian group by Problem “If the Quotient by the Center is Cyclic, then the Group is Abelian”.

Therefore, either $Z(G)=1$ or $G$ is abelian.

The post Group of Order $pq$ is Either Abelian or the Center is Trivial first appeared on Problems in Mathematics.

Suppose the order of a group $G$ is $p^2$, where $p$ is a prime number.
Show that

(a) the group $G$ is an abelian group, and

(b) the group $G$ is isomorphic to either $\Zmod{p^2}$ or $\Zmod{p} \times \Zmod{p}$ without using the fundamental theorem of abelian groups.

Hint.

Review the following problems.

1. The center of a p-group is not trivial (post 1)
2. If the quotient by the center is cyclic, then the group is abelian (post 2)

Proof.

(a) A group of order $p^2$ is abelian.

Since $G$ is a $p$-group, its center is not trivial (see post 1   for a proof.)

If the center $Z(G)=G$, then $G$ is abelian so assume that $Z(G)$ is a proper nontrivial subgroup. Then the center must have order $p$ and it follows that the order of the quotient $G/Z(G)$ is $p$, hence $G/Z(G)$ is a cyclic group.

Thus $G$ is abelian by the fact proved in post 2.

(b) The group $G$ is isomorphic to either $\Zmod{p^2}$ or $\Zmod{p} \times \Zmod{p}$

Let $x \in G$ be any nontrivial element of $G$. If $x \in G$ has order $p^2$, then $G=\langle x \rangle \cong \Zmod{p^2}$. If the order of $x$ is $p$, then take $y \in G \setminus \langle x \rangle$. Then the order of $y$ is also $p$.

Since the subgroup $\langle x, y \rangle$ generated by $x$ and $y$ is properly bigger than the subgroup $\langle x \rangle$, we must have $G=\langle x, y \rangle$.
We claim that $\langle x, y \rangle \cong \langle x \rangle \times \langle y \rangle$.

Define a map $f:\langle x \rangle \times \langle y \rangle \to \langle x, y \rangle$ by sending $(x^a, y^b)$ to $x^ay^b$. This is a group homomorphism because for any elements $(x^{a_1}, y^{b_1})$ and $(x^{a_2}, y^{b_2})$, we have
\begin{align*}
f\left( (x^{a_1}, y^{b_1})(x^{a_2}, y^{b_2}) \right) &=f\left (x^{a_1+a_2}, y^{b_1+b_2})\right) =x^{a_1+a_2} y^{b_1+b_2}\\
& =x^{a_1}y^{b_1}x^{a_2}y^{b_2} =f\left( (x^{a_1}, y^{b_1}) \right) f\left( (x^{a_2}, y^{b_2}) \right).
\end{align*}

Here we used the result of part (a) that $G$ is abelian in the third equality.

We claim that the homomorphism $f$ is injective.
If $f\left (x^a, y^b) \right)=1$, we have $x^a=y^b$ but since $y \not \in \langle x \rangle$ we must have $a=b=0$. Thus the kernel is trivial, hence $f$ is injective.

Since $\langle x \rangle \times \langle y \rangle \cong \Zmod{p} \times \Zmod{p}$ has order $p^2$ and $f$ is injective, the homomorphism must be surjective as well, hence it is an isomorphism.

The post A Group of Order the Square of a Prime is Abelian first appeared on Problems in Mathematics.

Let $Z(G)$ be the center of a group $G$.
Show that if $G/Z(G)$ is a cyclic group, then $G$ is abelian.

Steps.

1. Write $G/Z(G)=\langle \bar{g} \rangle$ for some $g \in G$.
2. Any element $x\in G$ can be written as $x=g^a z$ for some $z \in Z(G)$ and $a \in \Z$.
3. Using this expression, show that $xy=yx$ for any $x, y \in G$.

Proof.

Since the quotient group $G/Z(G)$ is cyclic, it is generated by one element.
Let $g \in G$ be an element such that $\bar{g}=gZ(G)$ is a generator of $G/Z(G)$. Namely, $\langle \bar g \rangle =G/ Z(G)$.

Then for any element $x \in G$, we have $\bar{x}\in G/Z(G)=\langle \bar g \rangle$ and hence $\bar{x}=\bar{g}^a$ for some $a \in \Z$.
It follows that any element of $G$ can be written as $x=g^a z$ for some $z \in Z(G)$ and $a \in \Z$.

Take any two elements $x, y \in G$ and write $x=g^a z$ and $y=g^b w$, where $a, b \in \Z$ and $z, w \in Z(G)$.

Then we claim that $xy=yx$. To see this we calculate as follows.
\begin{align*}
xy&= g^a z g^b w\\
&=g^a g^b z w && \text{since $z\in Z(G)$}\\
&= g^{a+b} wz && \text{since $w\in Z(G)$}\\
&=g^bg^awz\\
&=g^b w g^az&& \text{since $w\in Z(G)$}\\
&=yx.
\end{align*}

Since the product of any two elements of $G$ is commutative, we conclude that $G$ is abelian.

The post If the Quotient by the Center is Cyclic, then the Group is Abelian first appeared on Problems in Mathematics.