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 the group structure of the kernel of $\phi$.

Proof.

(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.

To show that $\phi$ is well-defined, we need show that the value of $\phi$ does not depends on the choice of representative $a$.
So suppose that $a+n\Z=a’+n\Z$ so that $a$ and $a’$ are two representatives for the same element.
This yields that $a-a’$ is divisible by $n$.

Now, $a+n\Z$ is mapped to $a+m\Z$ by $\phi$. On the other hand, $a’+n\Z$ is mapped to $a+m\Z$ by $\phi$.
Since $a-a’$ is divisible by $n$ and $m \mid n$, it follows that $a-a’$ is divisible by $m$.
This implies that $a+m\Z=a’+m\Z$.
This prove that $\phi$ does not depend on the choice of the representative, and hence $\phi$ is well-defined.

(b) Prove that $\phi$ is a group homomorphism.

Let $a+n\Z$, $b+n\Z$ be two elements in $\Zmod{n}$. Then we have
\begin{align*}
&\phi\left(\, (a+n\Z)+(b+n\Z) \,\right)\\
&=\phi\left(\, (a+b)+n\Z) \,\right) &&\text{by addition in $\Zmod{n}$}\\
&=(a+b)+m\Z &&\text{by definition of $\phi$}\\
&=(a+m\Z)+(b+m\Z)&&\text{by addition in $\Zmod{m}$}\\
&=\phi(a+n\Z)+\phi(b+n\Z) &&\text{by definition of $\phi$}.
\end{align*}

Hence $\phi$ is a group homomorphism.

(c) Prove that $\phi$ is surjective.

For any $c+m\Z \in \Zmod{m}$, we pick $c+n\Z\in \Zmod{n}$.
Then as $\phi(c+n\Z)=c+m\Z$, we see that $\phi$ is surjective.

(d) Determine the group structure of the kernel of $\phi$.

If $a+n\Z\in \ker(\phi)$, then we have $0+m\Z=\phi(a+n\Z)=a+m\Z$.
This implies that $m\mid a$.
On the other hand, if $m\mid a$, then $\phi(a+n\Z)=a+m\Z=0+m\Z$ and $a+n\Z\in \ker(\phi)$.

It follows that
\[\ker(\phi)=\{mk+n\Z \mid k=0, 1, \dots, l-1\},\] where $l$ is an integer such that $n=ml$.

Thus, $\ker(\phi)$ is a group of order $l$.
Since $\ker(\phi)$ is a subgroup of the cyclic group $\Zmod{n}$, we know that $\ker(\phi)$ is also cyclic.
Thus
\[\ker(\phi)\cong \Zmod{l}.\]

Another approach

Here is a more direct proof of this result.
Define a map $\psi:\Z\to \ker(\phi)$ by sending $k\in \Z$ to $mk+n\Z$.
It is straightforward to verify that $\psi$ is a surjective group homomorphism and the kernel of $\psi$ is $\ker(\psi)=l\Z$.
It follows from the first isomorphism theorem that
\[\Zmod{l}= \Z/\ker(\psi) \cong \im(\psi)=\ker(\phi). \] Click here if solved 125

The post Group Homomorphism from $\Z/n\Z$ to $\Z/m\Z$ When $m$ Divides $n$ first appeared on Problems in Mathematics.

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 the left cosets $G/P$ by left multiplication.
This induces a permutation representation homomorphism
\[\phi: G\to S_{G/P},\] where $S_{G/P}$ is a group of bijective maps (permutations) on $G/P$.

This homomorphism is defined by
\[\phi(g)(aP)=gaP\] for $g\in G$ and $aP\in G/P$.

Then by the first isomorphism theorem, we see that
\[G/\ker(\phi) \cong \im(\phi) < S_{G/P}.\] This implies that the order of $G/\ker(\phi)$ divides the order of $S_{G/P}$. Note that as $|G/P|=3$, we have $|S_{G/P}|=|S_3|=6$. Thus, we must have $4\mid |\ker{\phi}|$.

Also note that $\ker(\phi) < P$. To see this let $x\in \ker(\phi)$. Then we have \[xP=\phi(x)(P)=\id(P)=P.\] Here $\id$ is the identity map from $G/P$ to itself. Hence $x\in P$. It follows that $|\ker(\phi)|$ divides $|P|=8$.

Combining these restrictions, we see that $|\ker(\phi)|=4, 8$.
Being the kernel of a homomorphism, $\ker(\phi)$ is a normal subgroup of $G$.
Hence the group $G$ of order $24$ has a normal subgroup of order $4$ or $8$.

The post Every Group of Order 24 Has a Normal Subgroup of Order 4 or 8 first appeared on Problems in Mathematics.

Let $G, H, K$ be groups. Let $f:G\to K$ be a group homomorphism and let $\pi:G\to H$ be a surjective group homomorphism such that the kernel of $\pi$ is included in the kernel of $f$: $\ker(\pi) \subset \ker(f)$.

Define a map $\bar{f}:H\to K$ as follows.
For each $h\in H$, there exists $g\in G$ such that $\pi(g)=h$ since $\pi:G\to H$ is surjective.
Define $\bar{f}:H\to K$ by $\bar{f}(h)=f(g)$.

(a) Prove that the map $\bar{f}:H\to K$ is well-defined.

(b) Prove that $\bar{f}:H\to K$ is a group homomorphism.

Proof.

(a) Prove that the map $\bar{f}:H\to K$ is well-defined.

Let $h\in H$. Suppose that there are two elements $g, g’\in G$ such that $\pi(g)=h, \pi(g’)=h$.
Then we have
\begin{align*}
\pi(gg’^{-1})=\pi(g)\pi(g’)^{-1}=hh^{-1}=1
\end{align*}
since $\pi$ is a homomorphism.
Thus,
\[gg’^{-1}\in \ker(\pi) \subset \ker(f).\] It yields that $f(gg’^{-1})=1$.
It follows that
\begin{align*}
1=f(gg’^{-1})=f(g)f(g’)^{-1},
\end{align*}
and hence we have
\[f(g)=f(g’).\]

Therefore, the definition of $\bar{f}$ does not depend on the choice of elements $g\in G$ such that $\pi(g)=h$, hence it is well-defined.

(b) Prove that $\bar{f}:H\to K$ is a group homomorphism.

Our goal is to show that for any elements $h, h’\in H$, we have
\[\bar{f}(hh’)=\bar{f}(h)\bar{f}(h’).\]

Let $g, g’$ be elements in $G$ such that
\[\pi(g)=h \text{ and } \pi(g’)=h’.\] Then by definition of $\bar{f}$, we have
\[\bar{f}(h)=f(g) \text{ and } \bar{f}(h’)=f(g’) \tag{*}.\]

Since $\pi$ is a homomorphism, we have
\begin{align*}
hh’=\pi(g)\pi(g’)=\pi(gg’).
\end{align*}
By definition of $\bar{f}$, we have
\[\bar{f}(hh’)=f(gg’).\]

Since $f$ is a homomorphism, we obtain
\begin{align*}
\bar{f}(hh’)&=f(gg’)\\
&=f(g)f(g’)\\
&\stackrel{(*)}{=} \bar{f}(h)\bar{f}(h’).
\end{align*}
This proves that $\bar{f}$ is a group homomorphism.

The post A Group Homomorphism that Factors though Another Group first appeared on Problems in Mathematics.

Let $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer.
Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$.
Then prove that $H$ is a normal subgroup of $G$.

(Michigan State University, Abstract Algebra Qualifying Exam)

Proof.

Let $G/H$ be the set of left cosets of $H$.
Then the group $G$ acts on $G/H$ by the left multiplication.
This action induces the permutation representation homomorphism
\[\phi: G\to S_{G/H},\] where $S_{G/H}$ is the symmetric group on $G/H$.
For each $g\in G$, the map $\phi(g):G/H \to G/H$ is given by $x\mapsto gx$.

By the first isomorphism theorem, we have
\begin{align*}
G/\ker(\phi) \cong \im(\phi) < S_{G/H}.
\end{align*}
This implies the order $|G/\ker(\phi)|$ divides the order $|S_{G/H}|=p!$.

Since
\[|G/\ker(\phi)|=\frac{|G|}{|\ker(\phi)|}=\frac{p^n}{|\ker(\phi)|}\] and $p!$ contains only one factor of $p$, we must have either $|\ker(\phi)|=p^n$ or $|\ker(\phi)|=p^{n-1}$.

Note that if $g\in \ker(\phi)$, then $\phi(g)=\id:G/H \to G/H$.
This yields that $gH=H$, and hence $g\in H$.
As a result, we have $\ker(\phi) \subset H$.

Since the index of $H$ is $p$, the order of $H$ is $p^{n-1}$.
Thus we conclude that $|\ker(\phi)|=p^{n-1}$ and
\[\ker(\phi)=H.\]

Since every kernel of a group homomorphism is a normal subgroup, the subgroup $H=\ker(\phi)$ is a normal subgroup of $G$.

The post A Subgroup of Index a Prime $p$ of a Group of Order $p^n$ is Normal first appeared on Problems in Mathematics.

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 $K$.
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.

The post Group Homomorphisms From Group of Order 21 to Group of Order 49 first appeared on Problems in Mathematics.

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 $G$ 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 that
\[f(a)=1.\] Let $H=\langle a \rangle$ be the subgroup of $G$ generated by the element $a$.

We show that $G\cong \ker(f)\times H$.
To prove this isomorphism, it suffices to prove the following three conditions.

The first condition follows immediately since the group $G$ is abelian, hence all the subgroups of $G$ are normal.

To check condition 2, let $x\in \ker(f) \cap H$.
Then $x=a^n$ for some $n\in \Z$ and we have
\begin{align*}
0&=f(x) && \text{since $x \in \ker(f)$}\\
&=f(a^n)\\
&=nf(a) && \text{since $f$ is a homomorphism.}\\
&=n &&\text{since $f(a)=1$}.
\end{align*}
Thus, as a result we have $x=a^0=e$, and hence $\ker(f) \cap H=\{e\}$.
So condition 2 is met.

To prove condition 3, let $b$ be an arbitrary element in $G$.
Let $n=f(b) \in \Z$. Then we have
\[f(b)=n=f(a^n),\] and thus we have
\[f(ba^{-n})=0.\] It follows that $ba^{-n}\in \ker(f)$.
So there exists $z\in \ker(f)$ such that $ba^{-n}=z$.
Therefore we have
\begin{align*}
b=za^n\in \ker(f)H.
\end{align*}
This implies that $G=\ker(f)H$.

We have proved all the conditions, hence we obtain
\[G\cong \ker(f)\times H.\] Since $H$ is a cyclic group of infinite order, it is isomorphic to $\Z$.
(If $H$ has a finite order, then there exists a positive integer $n$ such that $a^n=e$. Then we have
\begin{align*}
0=f(e)=f(a^n)=nf(a)=n,
\end{align*}
and this contradicts the positivity of $n$.)

Combining these isomorphisms, we have
\[G\cong \ker(f)\times \Z,\] as required.

The post Surjective Group Homomorphism to $\Z$ and Direct Product of Abelian Groups first appeared on Problems in Mathematics.

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 monic.

Proof.

$(\implies)$ Injective implies monic

Suppose that $f: G \to G’$ is an injective group homomorphism.
We show that $f$ is monic.

So suppose that we have $fg_1=fg_2$, where $g_1:K\to G$ and $g_2:K \to G$ are group homomorphisms for some group $K$.
Then for any $x\in K$, we have
\[f(g_1(x))=f(g_2(x)).\] Since $f$ is injective, it follows that
\[g_1(x)=g_2(x)\] for any $x\in K$, and thus we obtain $g_1=g_2$. Thus $f$ is monic.

$(\impliedby)$ Monic implies injective

For the opposite implication, we prove the contrapositive statement. Namely, we prove that if $f$ is not injective, then $f$ is not monic.

Suppose that $f$ is not injective. Then the kernel $\ker(f)$ is a non-trivial subgroup of $G$.
We define the group homomorphism $g_1: \ker(f)\to G$ to be the identity map on $\ker(f)$. That is $g_1(x)=x$ for all $x\in \ker(f)$.
Also we define the group homomorphism $g_2:\ker(f)\to G$ by the formula $g_2(x)=e$ for all $x\in \ker(f)$, where $e$ is the identity element of $G$.
Since $\ker(f)$ is a nontrivial group, these two homomorphisms are distinct: $g_1\neq g_2$.

However, we have
\[fg_1=fg_2.\] In fact, we have for $x\in \ker(f)$
\[fg_1(x)=f(x)=e’,\] where $e’$ is the identity element of $G’$,
and
\[fg_2(x)=f(e)=e’.\] Thus, by definition, the homomorphism $f$ is not monic as required to complete the proof.

The post A Group Homomorphism is Injective if and only if Monic first appeared on Problems in Mathematics.

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 it induces the permutation representation $\rho: G \to S_n$, where $n=|G:H|$.
(Note that a permutation representation is a group homomorphism.)
Let $N=\ker \rho$ be the kernel of the homomorphism $\rho$. Then $N \triangleleft G$.

By the first isomorphism theorem, the quotient group $G/N$ is isomorphic to a subgroup of $S_n$. In particular, $G/N$ is a finite group, hence the index $[G:N]$ is finite.

Finally, we show that $N \subset H$.
For any $x \in N=\ker \rho$, we have $x(gH)=gH$ for any $g \in G$.
In particular we have $xH=H$, hence $x \in H$.

The post Subgroup of Finite Index Contains a Normal Subgroup of Finite Index first appeared on Problems in Mathematics.

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.

Proof.

$\Psi: G\to G$ is a group homomorphism

We first show that $\Psi: G\to G$ is a group homomorphism.
To see this, let $z, w \in G$. Then we have
\begin{align*}
\Psi(zw)=(zw)^p=z^pw^p=\Psi(z)\Psi(w).
\end{align*}
The second equality follows since $G$ is an abelian group.
Thus $\Psi$ is a group homomorphism.

$\Psi$ is surjective

To prove that $\Psi$ is surjective, let $z$ be an arbitrary element in $G$.
Then there exists nonnegative integer $n$ such that $z^n=1$.
Let $w \in \C$ be a $p$-th root of $z$, that is, $w$ is a solution of the equation $x^p-z=0$.
(By the fundamental theorem of algebra, such a solution exists.)

We check that $w \in G$ as follows.
We have
\begin{align*}
w^{p^{n+1}}=(w^p)^{p^n}=z^{p^n}=1.
\end{align*}
Therefore $w$ is a $p$-power root of $1$, hence $w\in G$.

It follows from
\begin{align*}
\Psi(w)=w^p=z
\end{align*}
that $\Psi$ is a surjective homomorphism.

$G$ is isomorphic to the proper quotient

Now by the first isomorphism theorem, we have an isomorphism
\[G/ \ker(\Psi) \cong \im(\Psi)=G.\] Since we have
\[\ker(\Psi)=\{z \in G \mid z^p=1\},\] the subgroup $\ker(\Psi)$ consists of $p$-th roots of unity.

There are $p$ $p$-th roots of unity in $\C$ (and hence in $G$), and hence the kernel $\ker(\Psi)$ is a nontrivial subgroup of $G$.
Hence $G/ \ker(\Psi)$ is a proper quotient, and thus $G$ is isomorphic to the proper quotient $G/ \ker(\Psi)$.

The post Group of $p$-Power Roots of 1 is Isomorphic to a Proper Quotient of Itself first appeared on Problems in Mathematics.