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!

  • 2
    $\begingroup$ 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

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.
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.