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I have to show that $$G=\{2n \mid n\in\mathbb{Z}\}$$ under addition is a group. I know for something to be a group it must satisfy $4$ things: identity, inverse, associativity and closure. I'm confused with what I'm supposed to do since I just have one element, $2n$, and I don't know what I'm suppose to do with the 'addition' part.

Thank you in advance!

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    $\begingroup$ You have infinitely many elements: ..., -4, -2, 0, 2, ... $\endgroup$ – k99731 Oct 8 '15 at 23:09
  • $\begingroup$ I know that. But how do I show that for like identity if ae=a=ea. $\endgroup$ – ematth7 Oct 8 '15 at 23:20
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The group operation is "ordinary addition", so for $a, b \in G$, $ab = a + b$. What you need to show is:

  • $0$ is even i.e., it's a member of $G$. ($0$ will be $e$, the unit of the group, and for $a \in G$, $ae = a + 0$.)
  • If $n$ is even, then $-n$ is even.
  • Associativity you get "for free", because ordinary addition is associative.
  • If $n, m$ are even, then $n+m$ is even.
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  • $\begingroup$ So why am I trying to show these things? $\endgroup$ – ematth7 Oct 8 '15 at 23:43
  • $\begingroup$ Um, so that you can show that $G$ forms a group with the group operation being ordinary addition. You do realize that $G$ is the set of even integers, yes? $\endgroup$ – BrianO Oct 8 '15 at 23:48
  • $\begingroup$ what would the difference be if it was multiplication? $\endgroup$ – ematth7 Oct 8 '15 at 23:52
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    $\begingroup$ It wouldn't be a group: it the operation were to be ordinary multiplication, the unit would have to be $1$, so inverses wouldn't exist – $1/2, 1/3, 1/4, \dots, 58/19, \dots$ aren't integers. The set of even integers is still "closed under multiplication", but the structure $(G, *)$ is only a monoid (set with an associative binary operation), not a group. $\endgroup$ – BrianO Oct 9 '15 at 0:06
  • $\begingroup$ Ahhh. So for identity could I write ae=a=ea & a*e=a+e=a, e=0. Or do I have to incorporate the 2n? $\endgroup$ – ematth7 Oct 9 '15 at 1:28

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