# Finite group of odd order with elements that are square

The problem I am struggling with is

"Find an example of a finite group G of order n where n is odd and each element of G is a square".

In my understanding this means

"Find a finite group with odd number of elements where each element can be expressed in ( )^2 form".

Am I understanding the question incorrectly?

I can't seem to find any finite group that fits this criteria.

Could anybody help me?

• Your understanding is correct. What's the simplest group you can think of? Does it work? – Patrick Stevens Jun 28 '18 at 6:00
• The trivial group satisfies this. – paul blart math cop Jun 28 '18 at 6:00
• Did you actually manage to find a group that did not fit the criteria? – Tobias Kildetoft Jun 28 '18 at 6:01
• – Angina Seng Jun 28 '18 at 6:12
• See also this question. – J.-E. Pin Jun 28 '18 at 6:17

Let $G$ be a finite group of odd order. Let $\phi: G \to G$ via $x \mapsto x^2$. Then we claim $\phi$ is an injection.

Let $|G| = n$. Suppose $x^2 = y^2$. $x = x^{n + 1}$. Then $n + 1$ is even, so let $n + 1 = 2k$. Similarly, $y = y^{2k}$. Then $x = x^{2k} = y^{2k} = y$. Thus $\phi$ is an injection. As $G$ is finite, it is therefore a surjection, so every element of $G$ is a square. Thus, all groups of odd order satisfy this.

• See my comment above... – J.-E. Pin Jun 28 '18 at 6:18

If $|G| = n$ then for any $g\in G$, $g^n = e$. So $g^{n+ 1} = g$ and $g= (g^{\frac {n+1}2})^2$ is a square.

So all groups with odd order satisfy this condition.

So you couldn't find any groups that failed the condition. You just didn't realize how they satisfied it.

....

(One thing to note. If the operation is "$*$" then a square is a number $m = x*x$. So if the operation is addition a "square" is a number $m = x + x$. So for, say, $\mathbb Z_5, +$ $0 = 0+0; 1 = 3+3; 2=1+1; 3=4+4; 4=2+2$ are all squares.