Let $G$ be a group. Suppose that the number of elements in $G$ of order $5$ is $28$.

Determine the number of distinct subgroups of $G$ of order $5$.

Solution.

Let $g$ be an element in $G$ of order $5$.
Then the subgroup $\langle g \rangle$ generated by $g$ is a cyclic group of order $5$.
That is, $\langle g \rangle=\{e, g, g^2, g^3, g^4\}$, where $e$ is the identity element in $G$.

Note that the order of each non-identity element in $\langle g \rangle$ is $5$.

Also, if $h$ is another element in $G$ of order $5$, then we have either $\langle g \rangle=\langle h \rangle$ or $\langle g \rangle \cap \langle h \rangle = \{e\}$.
This follows from the fact that the intersection $\langle g \rangle \cap \langle h \rangle$ is a subgroup of the order $5$ group $\langle g \rangle$, and thus the order of $\langle g \rangle \cap \langle h \rangle$ is either $5$ or $1$.

On the other hand, if $H$ is a subgroup of $G$ of order $5$, then every non-identity element in $H$ has order $5$.

These observations imply that each subgroup of order $5$ contains exactly $4$ elements of order $5$ and each element of order $5$ appears in exactly one of such subgroups.

As there are $28$ elements of order $5$, there are $28/4=7$ subgroups of order $5$.

The post If There are 28 Elements of Order 5, How Many Subgroups of Order 5? first appeared on Problems in Mathematics.

Let $G$ be a group and let $H_1, H_2$ be subgroups of $G$ such that $H_1 \not \subset H_2$ and $H_2 \not \subset H_1$.

(a) Prove that the union $H_1 \cup H_2$ is never a subgroup in $G$.

(b) Prove that a group cannot be written as the union of two proper subgroups.

Proof.

Prove that the union $H_1 \cup H_2$ is never a subgroup in $G$.

Seeking a contradiction, let us assume that the union $H_1 \cup H_2$ is a subgroup of $G$.
Since $H_1 \not \subset H_2$, there exists an element $a\in H_1$ such that $a\notin H_2$.
Similarly, as $H_2 \not \subset H_1$, there exists an element $b\in H_2$ such that $b\notin H_1$.

As we are assuming $H_1 \cup H_2$ is a group, we have $ab\in H_1 \cup H_2$.
It follows that either $ab \in H_1$ or $ab \in H_2$.

If $ab \in H_1$, then we have
\[b=a^{-1}(ab) \in H_1\] as both $a^{-1}$ and $ab$ are elements in the subgroup $H_1$.
This contradicts our choice of the element $b$.

Similarly, if $ab \in H_2$, we have
\[ a=(ab)b^{-1} \in H_2,\] which contradicts the choice of $a$.

In either case, we reached a contradiction.
Thus, we conclude that the union $H_1 \cup H_2$ is not a subgroup of $G$.

(b) Prove that a group cannot be written as the union of two proper subgroups.

This is a special case of part (a).

If a group $G$ is a union of two proper subgroup $H_1$ and $H_2$, then we must have $H_1 \not \subset H_2$ and $H_2 \not \subset H_1$, otherwise $G=H_1$ or $G=H_2$ and this is impossible as $H_1, H_2$ are proper subgroups.
Then $G=H_1\cup H_2$ is a subgroup of $G$, which is prohibited by part (a).

Thus, any group cannot be a union of proper subgroups.

The post Union of Two Subgroups is Not a Group first appeared on Problems in Mathematics.

Let $G$ be a finite group and let $N$ be a normal subgroup of $G$.
Suppose that the order $n$ of $N$ is relatively prime to the index $|G:N|=m$.

(a) Prove that $N=\{a\in G \mid a^n=e\}$.

(b) Prove that $N=\{b^m \mid b\in G\}$.

Proof.

Note that as $n$ and $m$ are relatively prime integers, there exits $s, t\in \Z$ such that
\[sn+tm=1. \tag{*}\] Also, note that as the order of the group $G/N$ is $|G/N|=|G:N|=m$, we have
\[g^mN=(gN)^m=N\] for any $g \in G$ by Lagrange’ theorem, and thus
\[g^m\in N \tag{**}.\]

(a) Prove that $N=\{a\in G \mid a^n=e\}$.

Suppose $a\in \{a\in G \mid a^n=e\}$. Then we have $a^n=e$.
It follows that
\begin{align*}
a \stackrel{(*)}{=} a^{sn+tm}= a^{sn}a^{tm}=a^{tm}=(a^t)^m\in N
\end{align*}
by (**).
This proves that $\{a\in G \mid a^n=e\} \subset N$.

On the other hand, if $a\in N$, then we have $a^n=e$ as $n$ is the order of the group $N$.
Hence $N\subset \{a\in G \mid a^n=e\}$.

Putting together these inclusions yields that $N=\{a\in G \mid a^n=e\}$ as required.

(b) Prove that $N=\{b^m \mid b\in G\}$.

Let $b^m \in \{b^m \mid b\in G\}$. Then by (**), we know that $b^m\in N$.
Thus, we have $\{b^m \mid b\in G\}\subset N$.

On the other hand, let $a\in N$. Then we have $a^n=e$ as $n=|N|$.
Hence it follows that
\begin{align*}
a \stackrel{(*)}{=} a^{sn+tm}= a^{sn}a^{tm}=a^{tm}=b^m,
\end{align*}
where we put $b:=a^t$.
This implies that $a\in \{b^m \mid b\in G\}$, and hence we have $N \subset \{b^m \mid b\in G\}$.

So we see that $N=\{b^m \mid b\in G\}$ by these two inclusions.

The post Normal Subgroup Whose Order is Relatively Prime to Its Index first appeared on Problems in Mathematics.

Let $G$ be a finite group. Let $S$ be the set of elements $g$ such that $g^5=e$, where $e$ is the identity element in the group $G$.

Prove that the number of elements in $S$ is odd.

Proof.

Let $g\neq e$ be an element in the group $G$ such that $g^5=e$.
As $5$ is a prime number, this yields that the order of $g$ is $5$.

Consider the subgroup $\langle g \rangle$ generated by $g$.
As the order of $g$ is $5$, the order of the subgroup $\langle g \rangle$ is $5$.

If $h\neq e$ is another element in $G$ such that $h^5=e$, then we have either $\langle g \rangle=\langle h \rangle$ or $\langle g \rangle \cap \langle h \rangle=\{e\}$ as the intersection of these two subgroups is a subgroup of $\langle g \rangle$.

It follows that $S$ is the union of subgroups of order $5$ that intersect only at the identity element $e$.
Thus the number of elements in $S$ are $4n+1$ for some nonnegative integer $n$.

The post The Number of Elements Satisfying $g^5=e$ in a Finite Group is Odd first appeared on Problems in Mathematics.

Let $G$ be a nilpotent group and let $H$ be a proper subgroup of $G$.

Then prove that $H \subsetneq N_G(H)$, where $N_G(H)$ is the normalizer of $H$ in $G$.

Proof.

Note that we always have $H \subset N_G(H)$.
Hence our goal is to find an element in $N_G(H)$ that does not belong to $H$.

Since $G$ is a nilpotent group, it has a lower central series
\[ G=G^{0} \triangleright G^{1} \triangleright \cdots \triangleright G^{n}=\{e\},\] where $G=G^{0}$ and $G^{i}$ is defined by
\[G^i=[G^{i-1},G]=\langle [x,y]=xyx^{-1}y^{-1} \mid x \in G^{i-1}, y \in G \rangle\] successively, and $e$ is the identity element of $G$.

Since $H$ is a proper subgroup of $G$, there is an index $k$ such that
\[G^{k+1} \subset H \text{ but } G^{k} \nsubseteq H.\]

Take any $x\in G^{k} \setminus H$.
We claim that $x \in N_G(H)$.

For any $y\in H$, it follows from the definition of $G^{k+1}$ that
\[ [x,y] \in G^{k+1} \subset H.\] Hence $xyx^{-1}y^{-1}\in H$.
Since $y\in H$, we see that $xyx^{-1}\in H$.
As this is true for any $y\in H$, we conclude that $x\in N_G(H)$.
The claim is proved.

Since $x$ does not belong to $H$, we conclude that $H \subsetneq N_G(H)$.

The post The Normalizer of a Proper Subgroup of a Nilpotent Group is Strictly Bigger first appeared on Problems in Mathematics.

Let $G$ be an abelian group and let $H$ be the subset of $G$ consisting of all elements of $G$ of finite order. That is,
\[H=\{ a\in G \mid \text{the order of $a$ is finite}\}.\]

Prove that $H$ is a subgroup of $G$.

Proof.

Note that the identity element $e$ of $G$ has order $1$, hence $e\in H$ and $H$ is not an empty set.

To show that $H$ is a subgroup of $G$, we need to show that $H$ is closed under multiplications and inverses.

Let $a, b\in H$.
By definition of $H$, the orders of $a, b$ are finite.
So let $m, n \in \N$ be the orders of $a, b$, respectively:
We have
\[a^m=e \text{ and } b^n=e.\]

Then we have
\begin{align*}
(ab)^{mn}&=a^{mn}b^{mn} && \text{since $G$ is abelian}\\
&=(a^m)^n(b^n)^m\\
&=e^ne^m=e.
\end{align*}

This implies that the order of $ab$ is at most $mn$, hence the order of $ab$ is finite.
Thus $ab\in H$ for any $a, b\in H$.

Next, consider any $a\in H$. We want to show that the inverse $a^{-1}$ also lies in $H$.
Let $m \in \N$ be the order of $a$.
Then we have
\begin{align*}
(a^{-1})^m=(a^m)^{-1}=e^{-1}=e.
\end{align*}
This implies that the order of $a^{-1}$ is also finite, and hence $a^{-1}\in H$.

Therefore we have proved that $H$ is closed under multiplications and inverses.
Hence $H$ is a subgroup of $G$.

The post Elements of Finite Order of an Abelian Group form a Subgroup first appeared on Problems in Mathematics.

Give an example of two groups $G$ and $H$ and a subgroup $K$ of the direct product $G\times H$ such that $K$ cannot be written as $K=G_1\times H_1$, where $G_1$ and $H_1$ are subgroups of $G$ and $H$, respectively.

Solution.

Let $G$ be any nontrivial group, and let $G=H$.
(For example, you may take $G=H=\Zmod{2}$.)

Then consider the subset $K$ in the direct product given by
\[K:=\{(g,g) \mid g\in G\} \subset G\times G.\]

We claim that $K$ is a subgroup of $G\times G$.
In fact, we have
\begin{align*}
(g,g)(h,h)=(gh,gh)\in K \text{ and }\\
(g,g)^{-1}=(g^{-1}, g^{-1})\in K
\end{align*}
for any $g, h\in G$.
Thus, $K$ is closed under multiplications and inverses, and hence $K$ is a subgroup of $G\times G$.

Now we show that $K$ is not of the form $G_1\times H_1$ for some subgroups $G_1, H_1$ of $G$.
Assume on the contrary $K=G_1\times H_1$ for some subgroups $G_1, H_1$ of $G$.

Since $G$ is a nontrivial group, there is a nonidentity element $x\in G$.
So $(x,x)\in K$ and $K$ is not the trivial group.
Thus, both $G_1$ and $H_1$ cannot be the trivial group.

Without loss of generality, assume that $G_1$ is nontrivial.
Then $G_1$ contains a nonidentity element $y$.

Since the identity element $e$ is contained in all subgroups, we have
\[(y,e)\in G_1\times H_1.\] However, this element cannot be in $K$ since $y\neq e$, a contradiction.

Hence $K$ is not of the form $G_1\times H_1$.

The post Example of Two Groups and a Subgroup of the Direct Product that is Not of the Form of Direct Product first appeared on Problems in Mathematics.

Problem 448

Let $G$ be a group. Let $H$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$.
The product of $H$ and $N$ is defined to be the subset
\[H\cdot N=\{hn\in G\mid h \in H, n\in N\}.\] Prove that the product $H\cdot N$ is a subgroup of $G$.

Definition.

A subgroup $N$ of a group $G$ is called a normal subgroup if for any $g\in G$ and $n\in N$, we have
\[gng^{-1}\in N.\]  

Proof.

We prove that the product $H\cdot N$ is closed under products and inverses.

Let $h_1n_1$ and $h_2n_2$ be elements in $H\cdot N$, where $h_1, h_2\in H$ and $n_1, n_2\in N$.
Let $e$ be the identity element in $G$.
We have
\begin{align*}
(h_1n_1)(h_2n_2)&=h_1en_1h_2n_2\\
&=h_1(h_2h_2^{-1})n_1h_2n_2 && \text{since $h_2h_2^{-1}=e$}\\
&=(h_1h_2)(h_2^{-1}n_1h_2n_2). \tag{*}
\end{align*}
Since $H$ is a subgroup, the element $h_1h_2$ is in $H$.
Also, since $N$ is a normal subgroup, we have $h_2^{-1}n_1h_2$ is in $N$. Hence
\[h_2^{-1}n_1h_2n_2=(h_2^{-1}n_1h_2)n_2\in N.\] It follows from (*) that the product
\[(h_1n_1)(h_2n_2)=(h_1h_2)(h_2^{-1}n_1h_2n_2)\in H\cdot N.\] Therefore, the product $H\cdot N$ is closed under products.

Next, let $hn$ be any element in $H\cdot N$, where $h\in H$ and $n\in N$.
Then we have
\begin{align*}
(hn)^{-1}&=n^{-1}h^{-1}\\
&=en^{-1}h^{-1}\\
&=(h^{-1}h)n^{-1}h^{-1} &&\text{since $h^{-1}h=e$}\\
&=h^{-1}(hn^{-1}h^{-1}).
\end{align*}
Since $N$ is a normal subgroup, we have $hn^{-1}h^{-1}\in N$, and hence
\[(hn)^{-1}=h^{-1}(hn^{-1}h^{-1})\in H\cdot N.\] Thus, the product $H\cdot N$ is closed under inverses.

This completes the proof that the product $H\cdot N$ is a subgroup of $G$.

The post The Product of a Subgroup and a Normal Subgroup is a Subgroup first appeared on Problems in Mathematics.

Let $G$ be an abelian group and let $N$ be a normal subgroup of $G$.
Then prove that the quotient group $G/N$ is also an abelian group.

Proof.

Each element of $G/N$ is a coset $aN$ for some $a\in G$.
Let $aN, bN$ be arbitrary elements of $G/N$, where $a, b\in G$.

Then we have
\begin{align*}
(aN)(bN)&=(ab)N \\
&=(ba)N && \text{since $G$ is abelian}\\
&=(bN)(aN).
\end{align*}
Here the first and the third equality is the definition of the group operation of $G/N$.

Remark

Since $N$ is a normal subgroup of $G$, the set of left cosets $G/H$ becomes a group with group operation
\[(aN)(bN)=(ab)N\] for any $a, b\in G$.

Related Question.

As an application, try the following problem.

The proof of this problem is given in the post ↴
If quotient $G/H$ is abelian group and $H < K \triangleleft G$, then $G/K$ is abelian.

The post Quotient Group of Abelian Group is Abelian first appeared on Problems in Mathematics.

Let $G=\GL(n, \R)$ be the general linear group of degree $n$, that is, the group of all $n\times n$ invertible matrices.
Consider the subset of $G$ defined by
\[\SL(n, \R)=\{X\in \GL(n,\R) \mid \det(X)=1\}.\] Prove that $\SL(n, \R)$ is a subgroup of $G$. Furthermore, prove that $\SL(n,\R)$ is a normal subgroup of $G$.
The subgroup $\SL(n,\R)$ is called special linear group

Hint.

We are going to use the following facts from linear algebra about the determinant of a matrix.
For any $n\times n$ matrices $A, B$, we have
\begin{align*}
\det(AB)&=\det(A)\det(B) \text{ and }\\
\det(A^{-1})&=\det(A)^{-1}
\end{align*}
if $A$ is invertible.

We give two proofs.
The first one proves that $\SL(n,\R)$ is a normal subgroup of $\GL(n,\R)$ by directly verifying the defining property.

The second proof uses a fact about group homomorphism. If you are familiar with group homomorphism, the second proof is concise and nice.

Proof 1.

The special linear group $\SL(n,\R)$ is a subgroup.

Let $X, Y\in \SL(n,\R)$ be arbitrary elements. We have
\[\det(X)=\det(Y)=1\] by definition of $\SL(n,\R)$.

Then we obtain
\begin{align*}
\det(XY)=\det(X)\det(Y)=1,
\end{align*}
and hence $XY$ is in $\SL(n,\R)$.

Also, we have
\[\det(X^{-1})=\det(X)^{-1}=1,\] and it follows that $X^{-1}$ is in $\SL(n,\R)$.
Thus, $\SL(n,\R)$ is a subgroup of $G$.

The special linear group $\SL(n,\R)$ is normal.

To prove that $\SL(n,\R)$ is a normal subgroup of $G$, let $X\in \SL(n,\R)$ and let $P\in G$.
Then we have
\begin{align*}
\det(PXP^{-1})=\det(P)\det(X)\det(P)^{-1}=\det(X)=1,
\end{align*}
and hence the conjugate $PXP^{-1}$ is in $\SL(n,\R)$.
Therefore, $\SL(n,\R)$ is a normal subgroup of $G$.

Proof 2.

Let $\R^*$ be the multiplicative group of nonzero real numbers.

Let $\phi:\GL(n,\R) \to \R^*$ be the map given by
\[\phi(X)=\det(X),\] for each $X\in \GL(n,\R)$.

Note that this is well-defined since $\det(X)\neq 0$ for $X\in \GL(n,\R)$.

By the property of the determinant, we know that
\[\det(XY)=\det(X)\det(Y)\] for any $X, Y \in \GL(n,\R)$.

This implies that the map $\phi$ is a group homomorphism.

Then the kernel of $\phi$ is given by
\begin{align*}
\ker(\phi)=\{X \in \GL(n,\R) \mid \phi(X)=\det(X)=1\}=\SL(n, \R).
\end{align*}

As the kernel of the group homomorphism $\phi:\GL(n,\R) \to \R^*$ is alway a normal subgroup of $\GL(n,\R)$, we conclude that $\SL(n, \R)$ is a normal subgroup of $\GL(n,\R)$.

The post Special Linear Group is a Normal Subgroup of General Linear Group first appeared on Problems in Mathematics.