Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$.

(a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$.

(b) Determine the index $[G : S]$.

(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$.

Proof.

(a) Prove that $S=\{x^2\mid x\in G\}$ is a subgroup of $G$.

Consider the map $\phi:G \to G$ defined by $\phi(x)=x^2$ for $x\in G$.
Then $\phi$ is a group homomorphism. In fact, for any $x, y \in G$, we have
\begin{align*}
\phi(xy)=(xy)^2=x^2y^2=\phi(x)\phi(y)
\end{align*}
as $G$ is an abelian group.

By definition of $\phi$, the image is $\im(\phi)=S$.
Since the image of a group homomorphism is a group, we conclude that $S$ is a subgroup of $G$.

(b) Determine the index $[G : S]$.

By the first isomorphism theorem, we have
\[G/\ker(\phi)\cong S.\]

If $x\in \ker(\phi)$, then $x^2=1$.
It follows that $(x-1)(x+1)=0$ in $\Zmod{p}$.
Since $\Zmod{p}$ is an integral domain, it follows that $x=\pm 1$ and $\ker(\phi)=\{\pm 1\}$.

Thus, $|S|=|G/\ker(\phi)|=(p-1)/2$ and hence the index is
\[[G:S]=|G|/|S|=2.\]

(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$.

Since $-1\notin S$ and $[G:S]=2$, we have the decomposition
\[G=S\sqcup (-S).\] Suppose that an element $a$ in $G$ is not in $S$.

Then, we have $a\in -S$.
Thus, there exists $b\in S$ such that $a=-b$.
It follows that $-a=b\in S$. Therefore, we have either $a\in S$ or $-a\in S$.

The post The Set of Square Elements in the Multiplicative Group $(\Zmod{p})^*$ first appeared on Problems in Mathematics.

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 124

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

Let $(\Q, +)$ be the additive group of rational numbers and let $(\Q_{ > 0}, \times)$ be the multiplicative group of positive rational numbers.

Prove that $(\Q, +)$ and $(\Q_{ > 0}, \times)$ are not isomorphic as groups.

Proof.

Suppose, towards a contradiction, that there is a group isomorphism
\[\phi:(\Q, +) \to (\Q_{ > 0}, \times).\]

Then since $\phi$ is in particular surjective, there exists $r\in \Q$ such that $\phi(r)=2$.
As $r$ is a rational number, so is $r/2$.

It follows that we have
\begin{align*}
2&=\phi(r)=\phi\left(\, \frac{r}{2}+\frac{r}{2} \,\right)\\
&=\phi\left(\, \frac{r}{2} \,\right)\cdot\phi\left(\, \frac{r}{2} \,\right) &&\text{ because $\phi$ is a homomorphism}\\
&=\phi\left(\, \frac{r}{2} \,\right)^2.
\end{align*}

It yields that
\[\phi\left(\, \frac{r}{2} \,\right)=\pm \sqrt{2}.\]

However, this is a contradiction since $\phi\left(\, \frac{r}{2} \,\right)$ must be a positive rational number, yet $\sqrt{2}$ is not a rational number.

We conclude that there is no such group isomorphism, and hence the groups $(\Q, +)$ and $(\Q_{ > 0}\times)$ are not isomorphic as groups.

The post The Additive Group of Rational Numbers and The Multiplicative Group of Positive Rational Numbers are Not Isomorphic 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.

Prove that the symmetric group $S_n$, $n\geq 3$ is a semi-direct product of the alternating group $A_n$ and the subgroup $\langle(1,2) \rangle$ generated by the element $(1,2)$.
 

Definition (Semi-Direct Product).

Internal Semi-Direct-Product

Recall that a group $G$ is said to be an (internal) semi-direct product of subgroups $H$ and $K$ if the following conditions hold.

1. $H$ is a normal subgroup of $G$.
2. $H\cap K=\{e\}$, where $e$ is the identity element in $G$.
3. $G=HK$.

In this case, we denote the group by $G=H\rtimes K$.

External Semi-Direct Product

If $G$ is an internal semi-direct product of $H$ and $K$, it is an external semi-direct product defined by the homomorphism $\phi:K \to \Aut(H)$ given by mapping $k\in K$ to the automorphism of left conjugation by $k$ on $H$.
That is, $G \cong H \rtimes_{\phi} K$.

Proof.

Recall that each element of the symmetric group $S_n$ can be written as a product of transpositions (permutations which exchanges only two elements).
This defines a group homomorphism $\operatorname{sgn}:S_n\to \{\pm1\}$ that maps each element of $S_n$ that is a product of even number of transpositions to $1$, and maps each element of $S_n$ that is a product of odd number of transpositions to $-1$.

The alternating group $A_n$ is defined to be the kernel of the homomorphism $\operatorname{sgn}:S_n \to \{\pm1\}$:
\[A_n:=\ker(\operatorname{sgn}).\]

As it is the kernel, the alternating group $A_n$ is a normal subgroup of $S_n$.
Also by first isomorphism theorem, we have
\[S_n/A_n\cong \{\pm1\},\] and it yields that
\[|A_n|=\frac{|S_n|}{|\{\pm1\}|}=\frac{n!}{2}.\]

Since $\operatorname{sgn}\left(\,(1,2) \,\right)=-1$, the intersection of $A_n$ and $\langle(1,2)\rangle$ is trivial:
\[A_n \cap \langle(1,2) \rangle=\{e\}.\]

Let $H=A_n$ and $K=\langle(1,2) \rangle$.
Then we have
\begin{align*}
|HK|=\frac{|H|\cdot |K|}{|H\cap K|}=|H|\cdot | K|=\frac{n!}{2}\cdot 2=n!.
\end{align*}
Since $HK < S_n$ and both groups have order $n!$, we have $S_n=HK$.

In summary we have observed that $H=A_n$ and $K=\langle(1,2) \rangle$ satisfies the conditions for a semi-direct product of $G=S_n$.
Hence
\[S_n=A_n\rtimes \langle(1,2) \rangle.\]

As an external semi-direct product, it is given by
\[S_n \cong A_n\rtimes_{\phi} \langle(1,2) \rangle,\] where $\phi: \langle(1,2) \rangle \to \Aut(A_n)$ is given by
\[\phi\left(\, (1,2) \,\right)(x)=(1,2)x(1,2)^{-1}.\] Click here if solved 31

The post The Symmetric Group is a Semi-Direct Product of the Alternating Group and a Subgroup $\langle(1,2) \rangle$ first appeared on Problems in Mathematics.

Let $G$ and $H$ be groups and let $\phi: G \to H$ be a group homomorphism.
Suppose that $f:G\to H$ is bijective.
Then there exists a map $\psi:H\to G$ such that
\[\psi \circ \phi=\id_G \text{ and } \phi \circ \psi=\id_H.\] Then prove that $\psi:H \to G$ is also a group homomorphism.

Proof.

Let $a, b$ be arbitrary elements of the group $H$.
To prove $\psi: H \to G$ is a group homomorphism, we need
\[\psi(ab)=\psi(a)\psi(b).\]

We compute
\begin{align*}
&\phi\left(\, \psi(a)\psi(b) \,\right)\\
&=\phi\left(\, \psi(a) \,\right) \phi\left(\, \psi(b) \,\right) && \text{since $\phi$ is a group homomorphism}\\
&=ab && \text{since $\phi\circ \psi=\id_H$}\\
&=\phi \left(\, \psi (ab) \,\right) && \text{since $\phi\circ \psi=\id_H$}.\\
\end{align*}

Since $\phi$ is injective, it yields that
\[\psi(ab)=\psi(a)\psi(b),\] and thus $\psi:H\to G$ is a group homomorphism.

What’s an Isomorphism?

A bijective group homomorphism $\phi:G \to H$ is called isomorphism.

The above problem guarantees that the inverse map of an isomorphism is again a homomorphism, and hence isomorphism.

The post Inverse Map of a Bijective Homomorphism is a Group Homomorphism first appeared on Problems in Mathematics.

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 if
\[\phi(ab)=\phi(a)\phi(b)\] for any elements $a, b\in G$.

Proof.

Let $e, e’$ be the identity elements of $G, G’$, respectively.
First we claim that
\[\phi(e)=e’.\] In fact, we have
\begin{align*}
\phi(e)&=\phi(ee)=\phi(e)\phi(e) \tag{*}
\end{align*}
since $\phi$ is a group homomorphism.
Thus, multiplying by $\phi(e)^{-1}$ on the left, we obtain
\begin{align*}
&e’=\phi(e)^{-1}\phi(e)\\
&=\phi(e)^{-1}\phi(e)\phi(e) && \text{by (*)}\\
&=e’\phi(e)=\phi(e).
\end{align*}
Hence the claim is proved.

Then we have
\begin{align*}
&e’=\phi(e) && \text{by claim}\\
&=\phi(gg^{-1})\\
&=\phi(g)\phi(g^{-1}) && \text{since $\phi$ is a group homomorphism}.
\end{align*}

It follows that we have
\begin{align*}
\phi(g)^{-1}&=\phi(g)^{-1}e’\\
&=\phi(g)^{-1}\phi(g)\phi(g^{-1})\\
&=e’\phi(g^{-1})\\
&=\phi(g^{-1}).
\end{align*}
This completes the proof.

The post Group Homomorphism Sends the Inverse Element to the Inverse Element first appeared on Problems in Mathematics.

Let $A=B=\Z$ be the additive group of integers.
Define a map $\phi: A\to B$ by sending $n$ to $2n$ for any integer $n\in A$.

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

(b) Prove that $\phi$ is injective.

(c) Prove that there does not exist a group homomorphism $\psi:B \to A$ such that $\psi \circ \phi=\id_A$.

Proof.

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

For any integers $m, n \in A$, we have
\begin{align*}
\phi(m+n)&=2(m+n)\\
&=2m+2n\\
&=\phi(m)+\phi(n).
\end{align*}
Thus, the map $\phi$ is a group homomorphism.

(b) Prove that $\phi$ is injective.

Suppose that we have
\[\phi(m)=\phi(n)\] for some integers $m, n\in A$.
This yields that we have $2m=2n$, and hence $m=n$.
So $\phi$ is injective.

Since $\phi$ is a group homomorphism, we can also prove the injectivity by showing that $\ker(\phi)=\{0\}$.
(For this, see the post “A Group Homomorphism is Injective if and only if the Kernel is Trivial“.)

Suppose that we have
\[\phi(m)=0.\] Then we have $2m=0$, and hence $m=0$.
It follows that the group homomorphism $\phi$ is injective.

(c) Prove that there does not exist a group homomorphism $\psi:B \to A$ such that $\psi \circ \phi=\id_A$.

Seeking a contradiction, assume that there exists a group homomorphism $\psi:B \to A$ such that $\psi \circ \phi =\id_A$.
Then we compute
\begin{align*}
&1=\id_A(1)=\psi \circ \phi(1)\\
&=\psi(2)=\psi(1+1)\\
&=\psi(1)+\psi(1) && \text{since $\psi$ is a group homomorphism}\\
&=2\psi(1).
\end{align*}
It yields that
\[\psi(1)=\frac{1}{2}.\] However note that $\psi(1)$ is an element in $A$, thus $\psi(1)$ is an integer.
Hence we got a contradiction, and we conclude that there is no such $\psi$.

The post Injective Group Homomorphism that does not have Inverse Homomorphism 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.