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.

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    $\begingroup$ 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. $\endgroup$ Commented Jul 7, 2023 at 16:56
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    $\begingroup$ 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. $\endgroup$ Commented Jul 7, 2023 at 16:57

1 Answer 1


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 $G$.


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