# Subgroup of $\mathbb{Z} \oplus \mathbb{Z}$ that is not a subring?

I just had a test, and one of the questions was to show that there is at least one subgroup of $\mathbb{Z} \oplus \mathbb{Z}$ that is not a subring.

I couldn't think of one and still can't, so I cheated and said $\{ (2k,2k) \colon \ k\in \mathbb{Z} \}$ is a subgroup but not a (unitary) subring... In this class we don't require a ring to have identity so it is wrong obviously.

Anyways can you please give me an example?

Thank you

• Just look at a few more subgroups. It shouldn't take you long to find a counterexample. Commented Mar 12, 2013 at 16:01
• You can use \oplus instead of \bigoplus, which you might prefer to save for things like $\bigoplus\limits_{p\text{ prime }}\Bbb Z/p\Bbb Z$, say.
– Pedro
Commented Mar 12, 2013 at 16:22
• @PeterTamaroff Thank you, I'll use that from now on. Commented Mar 12, 2013 at 16:51

Hint. Consider the subgroup $\{ (x, -x) : x \in \Bbb{Z} \}$.

• Ahh that makes sense. Thank you Commented Mar 12, 2013 at 16:28
• @Orlando, you're welcome! Commented Mar 12, 2013 at 16:33

Hint: Observe that $(1,2)·(1,2) = (1,4)$.

• $\mathbb Z_3$, by which I assume you mean $\mathbb Z/3$, is not a subgroup of $\mathbb Z$.
– Jim
Commented Mar 12, 2013 at 16:20
• @Orlando No. This, as Jim said, is not a subgroup of $ℤ$ or $ℤ × ℤ$. Instead consider $ℤ(1,2) = \{ (x,2x);\, x ∈ ℤ\}$ or what AndreasCaranti suggested. Check if they’re subgroups and whether they are multiplicatively closed or not. Commented Mar 12, 2013 at 16:27
• @K.Stm., thanks for the quote. Your example is fine, +1. Commented Mar 12, 2013 at 16:30
• @Orlando, subgroup means indeed same operation. Also note that $(x_1x_2, 2 (2x_1x_2))$ is not of the form $(x, 2x)$. i.e. second coordinate is twice the first one. Commented Mar 12, 2013 at 16:40
• First of all: $2(2x₁x₂) ≠ 2(x₁x₂)$ which was the requirement of being in that subgroup. (Put $x=x₁x₂$.) More concretely: $(1,2) ∈ ℤ(1,2)$ but $(1,2)·(1,2) = (1,4) \notin ℤ(1,2)$. Next, a subgroup $S$ of a bigger group $G$ must be a part of the bigger group in the sense that you need to have a structure preserving inclusion $S → G$. But since $1+1+1 = 0$ in $ℤ/3$ while $1+1+1 ≠ 0$ in $ℤ$, no such inclusion $ℤ/3 → ℤ$ exists. Commented Mar 12, 2013 at 16:43