finite odd order abelian group property

Question

question: let $|G|=odd $ where $G$ is a finite commutative group then to show every element of $G$ is a square.

ans 1> to show that $∀g∈G,∃g_1∈G,g=g_1^2$.

let $g \in G$ then $|g|$ $\big |$ $|G| \implies |g|=2n+1$ for some $n$

$e=g^{2n+1}=g^{2n}.g$ then if i take $g_1 =g^{-n}$ the result holds.

ans 2>

consider the homomorphism $\phi:G \rightarrow G$ definrd by $\phi(g)=g^2$ then $\phi$ is a homomorphism

now $ker{\phi} =\{ g\in G $ $\big|$ $ g^2=e \}$ then $ker (\phi) ={e}$ as no non trivial element of $G$ belong to $G$ as $2$ doesnot divide $|G|$

Hence $G \simeq \phi(G)$ and the result holds.

are both these answers correct?

More generally one can give an easy conceptual proof of $$\rm{\bf Theorem}\ \ \ g^n = 1,,\ (k,n)\mid i:\Rightarrow: g^i\ \ is\ a\ k'th\ power\ \ \ [{\bf\ Easy\ k'th\ Power\ Criterion}]$$ — Math Gems, Feb 15 '13 at 17:23

score 1 · Accepted Answer · answered Feb 15 '13 at 11:14

1

It can be generalized as follows:

Lemma: Let $G$ be a finite group of order $n$ and let $m$ be a positive integer such that $(m,n)=1$, then for all $g\in G$, there exists $x\in G$ with $g=x^m$.

The proof is routine and is the same as you did in ans 1.

answered Feb 15 '13 at 11:14

Mikasa

67,374

thanks for the generalization – jim Feb 15 '13 at 11:17
Also, the $x$ is unique. – Mikko Korhonen Feb 15 '13 at 11:34
More generally one can give an easy conceptual proof of $$\rm{\bf Theorem}\ \ \ g^n = 1,,\ (k,n)\mid i:\Rightarrow: g^i\ \ is\ a\ k'th\ power\ \ \ [{\bf\ Easy\ k'th\ Power\ Criterion}]$$ – Math Gems Feb 15 '13 at 17:19
@MathGems: Why didn't make your comment as an answer? ;-) – Mikasa Feb 15 '13 at 17:49

score 0 · Answer 2 · answered Feb 15 '13 at 06:54

0

Yes, both answers look fine.$\;$

answered Feb 15 '13 at 06:54

JSchlather

15,427

finite odd order abelian group property

2 Answers2