# 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? Oct 1, 2016 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. Oct 1, 2016 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. Oct 1, 2016 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 . Oct 1, 2016 at 18:19
• @PKStyles That's what I used Oct 1, 2016 at 19:20
• :How does $x=(x^{k+1})^2$ happen? Oct 1, 2016 at 20:53
• @PKStyles Since $n=2k+1$, $(x^{k+1})^2=x^{2k+2}=x^{n+1}=x^nx=x$ Oct 1, 2016 at 21:13
• :thanks for the response Oct 2, 2016 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. Oct 2, 2016 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 $m$).