A group has odd number of elements iff each element is a square, $a=b^2$.

Question

Let $G$ be a finite group having order $n$, prove that $n$ is odd if and only if for each $a\in G$, there is $b\in G$ such that $a=b^2$.

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I first assume $n$ is odd, then let $a$ be an element in $G$, since $a^n=e$, we have $$a=a^{n+1}=(a^{(n+1)/2})^2$$ which tells $a$ is the square of $a^{(n+1)/2}$.

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To prove the converse, I use contraposition, which I assume $n$ is even, then I want to find $a\in G$ such that it is not a square. My intuition is, "assume" we can always find an element $a\in G$ such that the smallest positive integer $k$ satisfy $$a^k=e$$ is $k=n$. Then it is not true that $a$ can be expressed as a square, indeed if $a=b^2$ for some $b\in G$, then the smallest $k$ satisfy $b^k=e$ will be $k=2n$, but this contradicts the smallest $k$ cannot exceed $n$.
My problem is, is the assumption I made always true? If it is true then it should be in the very beginning of the textbook, but I have forgotton it. Any help?

Source: Apostol Analytic Number Theory.

Answer 1

If I understand you right, then no. Just because $n$ is even does not mean there is an element $a$ where $n$ is the smallest power of $a$ that gives you $e$. For example, $G=C_2\times C_2$ has order $4$. And of course, $a^4=e$ for all $a\in G$. But also $a^2=e$ for all $a\in G$.

If $n$ is even, there is at least one non-identity element $a$ where $a^2=e$. So the map $G\to G$, where $x\mapsto x^2$ is not one-to-one. So the image of that map cannot be all of $G$.

Answer 2

Perhaps this helps.

Proposition Let $G$ be a finite group and $n$ a positive integer. Then the map $f: G \mapsto G$ defined by $f(g)=g^n$ is a bijection if and only if gcd$(|G|,n)=1$.

Proof (sketch) Bézout yields $1=k|G|+mn$, for some integers $k, m$. Then $g=g^{k|G|+mn}=g^{mn}=(g^m)^n$. Hence $f$ is surjective and since $G$ is finite it must be bijective. Conversely, assume gcd$(n,|G|)\neq 1$. Then we can find a prime $p$ with $p \mid n$ and $p \mid |G|$. By Cauchy's Theorem there is a non-trivial $g \in G$ with order$(g)=p$. Then $g^n=g^{p \cdot \frac{n}{p}}=1^\frac{n}{p}=1=1^n$. Since $f$ is injective this yields $g=1$, a contradiction.

Answer 3

No, that assumption is not true. That would mean that the group is cyclic, for which there are many counter-examples, such as $S_n, \, n \geq 3$.

Answer 4

Since the order of the group is divisble by 2, G has a 2-sylow subgroup. Consider an element of this 2-sylow subgroup that is of maximal order, i.e an element $l$ which has order $2^k$ with k max. Were it a square ($b^2=l$) then there would be an element of order $2^{(k+1)}$ contradicting the order maximality of $l$. A bit late to the party....