# Does this Lemma need commutativity?

I'm reading the book "Abstract Algebra, An Introduction" 3rd Edition by Hungerford and I'm trying to understand Lemma 9.4. The book states (and the book uses "+" for the group operation of abelian groups):

Let $$G$$ be an abelian group and $$a\in G$$ an element of finite order. Then $$a = a_1 + a_2 + · · · + a_t$$, with $$a_i\in G(p_i)$$, where $$p_1, ... , p_t$$ are the distinct positive primes that divide the order of $$a$$.

Here $$G(p_i)$$ are the elements of $$G$$ with order equal to a nonnegative power of $$p_i$$. Also, in case of doubt, the order of $$a$$ is the smallest integer $$k$$ such that $$ka=0$$.

The proof:

The proof is by induction on the number of distinct primes that divide the order of $$a$$. If $$|a|$$ (notation for order of $$a$$) is divisible only by the single prime $$p_1$$, then the order of $$a$$ is a power of $$p_1$$ and, hence, $$a\in G(p_1)$$. So the lemma is true in this case. Assume inductively that the lemma is true for all elements whose order is divisible by at most $$k - 1$$ distinct primes and that $$|a|$$ is divisible by the distinct primes $$p_1, ... , p_k$$. Then $$|a| = p_1^{r_1}\cdot ...\cdot p_k^{r_k}$$ with each $$r_i > 0$$. Let $$m = p_2^{r_2}\cdot ...\cdot p_k^{r_k}$$ and $$n = p_1^{r_1}$$, so that $$|a| = mn$$. Then $$(m,n) =1$$ and there are integers $$u, v$$ such that $$1 = mu + nv$$. Consequently, $$a= 1a = (mu + nv)a = mua + nva$$. But $$mua\in G(p_1)$$ because $$a$$ has order $$mn$$, and, hence; $$p_1^{r_1}\cdot (mua) = (nm )ua = u(mna) = u0 = 0$$. Similarly, $$m(nva) = 0$$ so that the order of $$nva$$ divides $$m$$, an integer with only $$k-1$$ distinct prime divisors. Therefore, by the induction assumption $$nva = a_2 + a_3 + · · · + a_k$$ with $$a_i\in G(p_i)$$. Let $$a_1 = mua$$; then $$a = mua + nva = a_1 + a_2 + · · · + a_k$$, with $$a_i\in G(p_i)$$.

My curiosity/confusion is where do we need the property that $$G$$ is abelian? I suspect it might have something to do with the last sentence because it's the only place with two distinct group elements.

• Yeah, you don't really need commutativity. Basically, you can choose the abelian subgroup generated by $a,$ so it extends to non-abelian groups, too. Commented Jul 7, 2023 at 16:56
• They are probably just proving for abelian groups so that they can use it to prove the "classification of finite abelian groups." But it is true for all groups. Commented Jul 7, 2023 at 16:57

You are right, you don't need the group to be abelian. If you are not sure, note that the abelian case also implies the general case. Consider the subgroup $$\langle a\rangle$$ generated by $$a$$. This is an abelian group, even if $$G$$ is not. So by the abelian case, the element $$a$$ has the required decomposition in $$\langle a\rangle$$, which is of course a decomposition in $$G$$.