Let $G$ be a group isomorphic to $Z_{n_1} \oplus Z_{n_2} \oplus \cdots\oplus Z_{n_k}$.

Question

Gallian's "Contemporary Abstract Algebra", Chapter 8 Problem 44:

Let $G$ be a group isomorphic to $Z_{n_1} \oplus Z_{n_2} \oplus \cdots \oplus Z_{n_k}$. Let $x$ be the product of all elements in $G$. Describe all possibilities for $x$.

I started with the simpler case of $G = Z_{n_1} \oplus Z_{n_2}$. If $\mathrm{gcd}(n_1, n_2) = 1$ then $G$ must be cyclic. The product of all elements would be $1+2+\ldots(n_1n_2-1)=(n_1n_2-1)(n_1n_2)/2 \mod{n_1n_2}$ in $Z_{n_1n_2}$. This seems unhelpful.

If $n_1$ and $n_2$ are not relatively prime, then adding up all the elements gives some $(x_1, x_2)$, where $x_1 = (n_1-1)n_1/2*n_2 \mod{n_1}$ and $x_2 = (n_2-1)n_2/2*n_1 \mod{n_2}$, again unhelpful.

Answer 1

In $Z_{n}$, the sum of all the elements is $\frac{n(n-1)}{2}$ mod $n$.

If $n$ is odd, this is $0$. If $n$ is even, this is $\frac n2$.

Let the sum of the elements of (abelian) $G$ be denoted $s(G)$. So $s(Z_n)=0$ if $n$ is odd, or $\frac n2$ if n is even.

In the direct product $H\oplus K$, the sum of all the elements will be $|H|s(K)+|K|s(H)$. Hence, for $G=Z_{n_1}\oplus\dots\oplus Z_{n_k}$, the sum of the elements will be $$\sum_{i=1}^k\frac{|G|}{n_i}s(n_i)$$

The $i$th term in the sum is $0$ unless $n_i$ is even AND $n_j$ is odd for all $j\neq i$.

So the possible values of the sum are:

• $0$
• $\frac {n_i}2$, if and only if $Z_{n_i}$ is the only even-order group in the sum.

Answer 2

Denote $n_1n_2=n$. If $n$ is even $1+2+...+n-1=(1+(n-1))+(2+(n-2))+...+((n/2-1)+(n/2+1))+n/2$.
So $(1+2+...+n-1)\equiv \frac{n}{2} $ mod $n$.
Similarly, if $n$ is odd, $(1+2+...+n-1)\equiv0$ mod $n$