Without using Lagrange's theorem, show that an abelian group of odd order cannot have an element of even order.

Let $G$ be an abelian group of odd order, say $n$. Then, by fundamental theorem of finite abelian groups, $G$ is a direct product of groups of prime power order. Let $$G=\mathbb Z_{{p_1}^{n_1}} \oplus \mathbb Z_{{p_2}^{n_2}}\oplus \cdots\oplus\mathbb Z_{{p_k}^{n_k}}.$$ Then $\vert G\vert=n={p_1}^{n_1}{p_2}^{n_2}\cdots{p_k}^{n_k}$. Now let $(g_1,g_2,...,g_k)\in G$. Then $\vert (g_1,g_2,...,g_k )\vert=\mathrm{lcm}(\vert g_1 \vert,\vert g_2 \vert,...,\vert g_k \vert)$.

From here how to move, i don't know but it is given in the hint to make use of fundamental theorem of cyclic groups (but i'm not getting how to apply it here).

• How do you get the decomposition into a direct sum of cyclic groups without Lagrange theorem? – egreg Oct 1 '16 at 17:23
• @egreg:Since G is abelian,then by fundamental theorem of finite abelian groups(which states that-every finite abelian group is the direct product of cyclic groups of prime power order) it can be represented as the direct product of cyclic groups. – P.Styles Oct 1 '16 at 17:36
• The point is that it does not make any sense to avoid using Lagrange's Theorem if you are going to use a more advanced theorem like the fundamental theorem of abelian groups. – Derek Holt Oct 1 '16 at 20:37

Assuming you can prove the fundamental theorem without using Lagrange (which I don't believe), a consequence of the theorem is that, for every $x\in G$, $x^n=1$, because $n$ is a multiple of the order of every cyclic component and, clearly, in a cyclic group $C$ of order $k$, we have $g^k=1$, for every $c\in C$.

The map $x\mapsto x^2$ is an endomorphism of $G$ and, since $n=2k+1$ is odd, we have $$x=(x^{k+1})^2$$ so the map is surjective. Thus the map is also injective, therefore $x^2=1$ implies $x=1$ and there's no element of order $2$.

• :I did'nt noticed this fact that proof of FTFAG makes use of Lagrange's theorem.But this exercise is from Gallian's algebra text.Whatever i mentioned in my post is directed through the hint in the very text.May be i'm wrong in understanding the hint.Hints are given as-use 1.Fundamental theorem of finite abelian groups.2.Order of elements in a direct product.3.Fundamental theorem of cyclic groups . – P.Styles Oct 1 '16 at 18:19
• @PKStyles That's what I used – egreg Oct 1 '16 at 19:20
• :How does $x=(x^{k+1})^2$ happen? – P.Styles Oct 1 '16 at 20:53
• @PKStyles Since $n=2k+1$, $(x^{k+1})^2=x^{2k+2}=x^{n+1}=x^nx=x$ – egreg Oct 1 '16 at 21:13
• :thanks for the response – P.Styles Oct 2 '16 at 4:41

Suppose $g\in G$ has even order, say $2m$, then $g^m$ has order $2$ so assume $g$ has order $2$, then we can partition $G$ into pairs, $\{h,k\}$ where $h=gk$ and $k=gh$.

This is of course impossible since $G$ has odd order, so we have the result by contradiction.

• Nice. This works even if $G$ is not abelian. – Arnaud D. Oct 2 '16 at 9:54

I agree with egreg that using the FTFGAG seems very much like overkill; and Robert Chamberlain's proof is very nice, but seems essentially to be re-proving Lagrange's theorem in this case. I wonder whether what Gallian has in mind isn't just to use the fact that multiplication by $g^n$ fixes the product of the elements of $G$, hence is trivial; so, if $g$ had even order $m$, then we would have that there existed integers $a$ and $b$ so that $g^{\gcd(m, n)} = g^{a m + b n} = (g^m)^a(g^n)^b = 1$. This is a contradiction, since $\gcd(m, n)$ is positive but strictly less than $m$ (it divides $m$, and is odd so it divides $n$, so it doesn't equal $m$).