In each case, I am asked to decide whether the indicated pair is a group or not. If so, prove it; if not, show which group axiom fails.

(a) $(\dfrac{1}{2}\mathbb{Z}, +)$ where $\dfrac{1}{2} \mathbb{Z} = \{\dfrac{n}{2} | n \in \mathbb{Z}\}$.

(b) $(\mathbb{R_{\geq 0}}, *)$ where $x * y = max\{x,y\}$.

As for (a), it appears that $\dfrac{1}{2}\mathbb{Z}$ is closed with respect to addition since given $x,y \in \dfrac{1}{2}\mathbb{Z}$, we have $x = \dfrac{n_1}{2}, y = \dfrac{n_2}{2}$, for some $n_1,n_2 \in \mathbb{Z}$ and so $x + y = \dfrac{n_1 + n_2}{2}$ and clearly $n_1 + n_2 \in \mathbb{Z}$ so $x+y \in \dfrac{1}{2} \mathbb{Z}$.

It's associative as well since $x,y,z \in \dfrac{1}{2}\mathbb{Z}$ is such that $x = \dfrac{n_1}{2}, y = \dfrac{n_2}{2}, z = \dfrac{n_3}{2} = \dfrac{n_1 + n_2 + n_3}{2}$ in either case.

There's an identity element since $0 \in \dfrac{1}{2}\mathbb{Z}$ because $ 0 = \dfrac{0}{n}$.

And there exists inverses since for any $x = \dfrac{n_0}{2} \in \dfrac{1}{2}\mathbb{Z}$ we have $x^{-1} = \dfrac{-n_0}{2}$ and $\dfrac{n_0 - n_0}{2} = 0 = e_{\dfrac{1}{2}\mathbb{Z}}$. So $(\dfrac{1}{2}\mathbb{Z},+)$ is a group.

(b) This is not a group since we do not have a $unique$ identity element since while $\forall x \in \mathbb{R_{\geq 0}}$ we have $x*x = x$, there is no unique $e$ such that $e*x = x$ that will not depend on $x$.

Is this reasoning correct?

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    $\begingroup$ For (b), $0$ has the property that $\max\{0,x\}=x$ for all $x\geq 0$. The problem comes with inverses. $\endgroup$ – Joe Johnson 126 Oct 19 '15 at 2:48
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    $\begingroup$ For (b), your reasoning is incorrect because of this: you are suggesting that since $x * x = x$ for every $x$, then suddenly every $x$ is an identity element. That's not true. For $x$ to be an identity element, it should be true that $x * y = y$ for every $y$. We don't have that since for each fixed nonzero $x$, taking any $y$ smaller than $x$ yields $x * y = x \neq y$. That means for each nonzero $x$, it's not an identity element. So there is only one identity element (do you know what it is?). I hope that helps. $\endgroup$ – layman Oct 19 '15 at 2:51
  • $\begingroup$ Thank you, I thought I may have hastily concluded that there is no identity element. I suppose the problem with inverses is that for any $r \neq 0 \in \mathbb{R_{\geq 0}}$ , $\sim \exists r^{-1}$ such that $ r * r^{-1} = 0$ since the lowest number we can get from x*y is either x or y, and neither is 0 in this case so we can't achieve $x*y =0$ for any $x,y \neq 0$. Is this why there is not inverse? $\endgroup$ – letsmakemuffinstogether Oct 19 '15 at 2:54
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    $\begingroup$ @letsmakemuffinstogether That sounds right to me! For every non-zero $x$, $x*y$ is at least $x$, and so it can't be $0$ since $x > 0$, and thus $x$ has no inverse. $\endgroup$ – layman Oct 19 '15 at 2:57
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    $\begingroup$ One thing to note is that while $xx=x$ for all $x$ doesn't imply that there doesn't exist an identity, or that the identity is not unique, it does imply that you can't have a group, since in a group you can cancel to conclude that $x=e$ for all $x$, which would be a contradiction if you have more than one element. $\endgroup$ – Joey Zou Oct 19 '15 at 3:00

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