Let $G$ be a group. Suppose that the order of nonidentity element of $G$ is $2$.
Then show that $G$ is an abelian group.

Proof.

Let $x$ and $y$ be elements of $G$. Then we have
\[1=(xy)^2=(xy)(xy).\]

Multiplying the equality by $yx$ from the right, we obtain
\begin{align*}
yx&=(xy)(xy)(yx)\\
&=xyxy^2x\\
&=xyx^2 \quad (\text{ since } y^2=1)\\
&=xy \quad (\text{ since } x^2=1).
\end{align*}
Thus we obtain $xy=yx$ for any elements $x, y \in G$. Thus the group $G$ is an abelian group.

The post If Every Nonidentity Element of a Group has Order 2, then it's an Abelian Group first appeared on Problems in Mathematics.

Let $G$ be a finite group of order $n$ and let $m$ be an integer that is relatively prime to $n=|G|$. Show that for any $a\in G$, there exists a unique element $b\in G$ such that
\[b^m=a.\]  

We give two proofs.

Proof 1.

Since $m$ and $n$ are relatively prime integers, there exists integers $s, t$ such that
\[sm+tn=1.\]

Then we have
\begin{align*}
a&=a^1=a^{sm+tn}=a^{sm}a^{tn} \tag{*}
\end{align*}
Note that since the order of the group $G$ is $n$, any element of $G$ raised by the power of $n$ is the identity element $e$ of $G$.
Thus we have
\[a^{tn}=(a^n)^t=e^t=e.\] Putting $b:=a^s$, we have from (*) that
\[a=b^me=b^m.\]

Now we show the uniqueness of such $b$. Suppose there is another $g’\in G$ such that
\[a=b’^m.\] Then we have
\begin{align*}
&\quad b^m=a=b’^m\\
&\Rightarrow b^{sm}=b’^{sm} \quad \text{ by taking $s$-th power}\\
&\Rightarrow b^{1-tn}=b’^{1-tn}\\
&\Rightarrow b(b^n)^t=b'(b’^n)^t\\
&\Rightarrow b=b’ \quad \text{ since } b^n=e=b’^n.
\end{align*}
Therefore, we have $b=b’$ and the element $b$ satisfying $a=b^m$ is unique.

Proof 2.

Consider a map $f$ from $G$ to $G$ itself defined by sending $g$ to $f(g)=g^m$.
We show that this map is injective.
Suppose that $f(g)=f(g’)$.

Since $m$ and $n$ are relatively prime integers, there exists integers $s, t$ such that
\[sm+tn=1.\]

We have
\begin{align*}
f(g)&=f(g’)\\
&\Rightarrow g^m=g’^m \\
&\Rightarrow g^{sm}=g’^{sm} \quad \text{ by taking $s$-th power}\\
&\Rightarrow g^{1-tn}=g’^{1-tn}\\
&\Rightarrow g(g^n)^t=g'(g’^n)^t\\
&\Rightarrow g=g’ \quad \text{ since } g^n=e=g’^n, n=|G|.
\end{align*}

Therefore the map $f$ is injective. Since $G$ is a finite set, it also follows that the map is bijective.
Thus for any $a \in G$, there is a unique $b \in G$ such that $f(b)=a$, namely, $b^m=a$.

The post Finite Group and a Unique Solution of an Equation first appeared on Problems in Mathematics.

Let $G$ be a finite group of odd order. Assume that $x \in G$ is not the identity element.

Show that $x$ is not conjugate to $x^{-1}$.
 

Proof.

Assume the contrary, that is, assume that there exists $g \in G$ such that $gx^{-1}g^{-1}=x$.
Then we have
\[xg=gx^{-1}. \tag{*}\] Then we compute
\begin{align*}
(xg)^2&=(gx^{-1})(xg)=g^2\\
(xg)^3&=(xg)(xg)^2=(xg)(g^2)=xg^3\\
(xg)^4&=(xg)^2(xg)^2=g^2g^2=g^4.
\end{align*}

In general, we have by induction
\begin{align*}
(xg)^k=
\begin{cases}
xg^k & \text{ if } k \text{ is odd}\\
g^k & \text{ if } k \text{ is even}. \tag{**}
\end{cases}
\end{align*}

Let $m$ be the order of $g$. Since $G$ is a finite group of odd order, $m$ is odd.
We have by (**)
\[(xg)^{m-1}g=g^{m-1}g=g^m=1.\] Thus we have $g^{-1}=(xg)^{m-1}$. Multiplying by $xg$ from the right, we have
\begin{align*}
g^{-1}xg &=(xg)^m\\
&=xg^m \text{ by (**)}\\
&=x.
\end{align*}

Thus we have $gx=xg$, and combining with (*) we obtain $gx=gx^{-1}$.
This implies that $x^2=1$, and since the order of $G$ is odd, this implies $x=1$.

This contradicts our choice of $x$.
Therefore $x$ cannot be conjugate to $x^{-1}$.

The post If a Group is of Odd Order, then Any Nonidentity Element is Not Conjugate to its Inverse first appeared on Problems in Mathematics.

Determine whether a group $G$ of the following order is simple or not.

(a) $|G|=100$.
(b) $|G|=200$.
 

Hint.

Use Sylow’s theorem and determine the number of $5$-Sylow subgroup of the group $G$.
Check out the post Sylow’s Theorem (summary) for a review of Sylow’s theorem.

Proof.

(a) When $|G|=100$.

The prime factorization of $100$ is $2^2\cdot 5^2$. Let us determine the number $n_5$ of $5$-Sylow subgroup of $G$.
By Sylow’s theorem, we know that $n_5 \equiv 1 \pmod{5}$ and $n_5$ divides $2^2$.
The only number satisfies both constraints is $n_5=1$. Thus there is only one $5$-Sylow subgroup of $G$. This implies that the $5$-Sylow subgroup is a normal subgroup of $G$.
Since the order of the $5$-Sylow subgroup is $25$, it is a proper normal subgroup. Thus, the group $G$ is not simple.

(b) When $|G|=200$

The prime factorization is $200=2^3\cdot 5^2$.
We again consider the number $n_5$ of $5$-Sylow subgroups of $G$.

Sylow’s theorem implies that $n_5 \equiv 1 \pmod{5}$ and $n_5$ divides $2^3$.
These two constraints has only one solution $n_5=1$.
Thus the group $G$ has a unique proper normal $5$-Sylow subgroup of order $25$. Hence $G$ is a simple group.

Similar problem

For an analogous problem, check out the post: If the order is an even perfect number, then a group is not simple
 

Comment.

This is the 100th problems in this blog.
To post 100 problems were not so simple.

The next goal is to archive the 200th problem.
This problem suggests that this goal is again not simple.
(Update: On 11/25/2016 I achieved the 200th problem: Maximize the dimension of the null space of $A-aI$.)

The post Are Groups of Order 100, 200 Simple? first appeared on Problems in Mathematics.

Let $p$ be a prime number. Suppose that the order of each element of a finite group $G$ is a power of $p$. Then prove that $G$ is a $p$-group. Namely, the order of $G$ is a power of $p$.

Hint.

You may use Sylow’s theorem.
For a review of Sylow’s theorem, please check out the post Sylow’s Theorem (summary).

Proof.

If $G$ is a trivial group, then the claim is trivial. So assume that $|G|>1$.

Seeking a contradiction, suppose that $|G|=p^nm$ for some $n,m \in \Z$ and $p$ and $m>1$ are relatively prime.
Let $l$ be a prime factor of $m$. Then by Sylow’s theorem, there exists a Sylow $l$-subgroup of $G$.
The order of a nontrivial element of this subgroup is divisible by the prime $l$ and this contradicts that each element has order power of $p$ since $l$ and $p$ are relatively prime.

Comment.

If we assume Sylow’s theorem, then the proof of this problem is straightforward.
How about proving it more directly (without using Sylow’s theorem)?

The post A Group with a Prime Power Order Elements Has Order a Power of the Prime. first appeared on Problems in Mathematics.