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

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 $$a=b^2$$.

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 $$a^{(n+1)/2}$$.
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.

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 $$G$$.

• The second paragraph really helps me, thank you! Feb 12, 2019 at 6:18
• I am only speaking of the group $C_2\times C_2$, not a ring. The set $\{e=(1,1), (1,-1), (-1,1),(-1,-1)\}$. And the group operation ins multiplication in each component. Feb 12, 2019 at 6:19

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.

• Yes, the point is Cauchy's Theorem, I should working on it, thank you! Feb 14, 2019 at 4:17

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

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....