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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

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    $\begingroup$ Just look at a few more subgroups. It shouldn't take you long to find a counterexample. $\endgroup$ Commented Mar 12, 2013 at 16:01
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    $\begingroup$ 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. $\endgroup$
    – Pedro
    Commented Mar 12, 2013 at 16:22
  • $\begingroup$ @PeterTamaroff Thank you, I'll use that from now on. $\endgroup$
    – Orlando
    Commented Mar 12, 2013 at 16:51

2 Answers 2

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

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  • $\begingroup$ Ahh that makes sense. Thank you $\endgroup$
    – Orlando
    Commented Mar 12, 2013 at 16:28
  • $\begingroup$ @Orlando, you're welcome! $\endgroup$ Commented Mar 12, 2013 at 16:33
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Hint: Observe that $(1,2)·(1,2) = (1,4)$.

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    $\begingroup$ $\mathbb Z_3$, by which I assume you mean $\mathbb Z/3$, is not a subgroup of $\mathbb Z$. $\endgroup$
    – Jim
    Commented Mar 12, 2013 at 16:20
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    $\begingroup$ @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. $\endgroup$
    – k.stm
    Commented Mar 12, 2013 at 16:27
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    $\begingroup$ @K.Stm., thanks for the quote. Your example is fine, +1. $\endgroup$ Commented Mar 12, 2013 at 16:30
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    $\begingroup$ @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. $\endgroup$ Commented Mar 12, 2013 at 16:40
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    $\begingroup$ 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. $\endgroup$
    – k.stm
    Commented Mar 12, 2013 at 16:43

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