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As an exercise in my intro to abstract algebra class, the lecturer posed the following problem:

Find all invertible multiplicative elements of $$ 1) \ \ \mathbb{Z} $$ $$ 2) \ \ \mathbb{Z}\times\mathbb{Z} $$

For the first I assume it would just be elements of the form $ \frac{1}{m} $ where $ m \in \mathbb{Z} $. I'm not sure what to do for the second. Please help.

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    $\begingroup$ Is $1/m$ an integer, for $m > 1$? $\endgroup$ – David Wheeler Mar 18 '15 at 7:47
  • $\begingroup$ If $|m|>1$, the rational number $\frac1m$ does not belong to $\Bbb Z$, so it cannot be an invertible element of $\Bbb Z$, and neither can $m$. $\endgroup$ – Brian M. Scott Mar 18 '15 at 7:48
  • $\begingroup$ I went back to undo edits I made earlier, thinking the problem was with the post. For some reason, MathJax wasn't working for me for about 36 hours (including no "edit window")... $\endgroup$ – colormegone Mar 19 '15 at 6:00
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Here's a "hint", since I think you don't understand quite what's going on and an answer won't help. The invertible multiplicative elements of any group/ring/whatever say $X$ are elements $u\in X$ such that there is $v\in X$ such that $uv=1$.

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  • $\begingroup$ see the above answer. If $u = 1$ and $v = -1$ then $uv = -1$ $\endgroup$ – user119264 Mar 18 '15 at 7:52
  • $\begingroup$ $u,v$ are allowed to be the same and the choice of $v$ depends on $u$. $\pm 1$ are the invertible elements of $\mathbb{Z}$ and they are their own inverses. $\endgroup$ – Moya Mar 18 '15 at 7:53
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Here is a "slightly" more general approach.

Define $N:\Bbb Z \to \Bbb N$, by $N(k) = |k|$ (the absolute value).

You can prove that $N(km) = N(k)N(m)$ (consider the $4$ cases:

1) $k,m \geq 0$

2) $k \geq 0, m < 0$

3) $k < 0, m \geq 0$

4) $k,m < 0$ separately).

An invertible (multiplicative) element $x \in \Bbb Z$ is one for which there is a $y \in \Bbb Z$, with $xy = 1$.

Thus: $N(x)N(y) = N(xy) = N(1) = 1$.

Now if $N(x) > 1$ (which is just saying $|x| > 1$), but $N(x)N(y) = 1$, then:

$1 = N(x)N(y) > N(y)$.

But the ONLY natural number less than $1$ is $0$, so that $y = 0$ and $N(0) = 0$, so:

$1 = N(x)N(y) = N(x)N(0) = N(x)0 = 0$, a contradiction.

If $N(x) = 0$, then $x = 0$, and $N(0)N(y) = 0$ (no matter what $y$ we pick), so $N(x)N(y) \neq 1$,

and so $0$ is not invertible.

That means if $x$ is invertible, we must have $N(x) = 1$. Which integers have $N(x) = 1$?

The only two are $1$ and $-1$, and we can check directly these two are invertible:

$(1)(1) = 1$ and $(-1)(-1) = 1$.

Now, see if you can extend this to $\Bbb Z \times \Bbb Z$ by showing that:

$(a,b)$ is invertible in $\Bbb Z \times \Bbb Z$ if and only if $a,b$ are both invertible in $\Bbb Z$ (recall that the multiplicative identity of $\Bbb Z \times \Bbb Z$ is $(1,1)$, and we multiply "coordinate-wise").


There is a method to my madness-for "more complicated" structures you will meet later, there are some similar $N$ functions, and you will get more mileage out of this approach.

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Idea is that : both $x, \dfrac{1}{x} \in \mathbb{Z}$, and this occurs when $x = \pm 1$.

$1)$ $1,-1$.

$2)$ $(1,1), (1,-1), (-1,1), (-1,-1)$.

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  • $\begingroup$ Can you please give an explanation as to why you made this choice? $\endgroup$ – user119264 Mar 18 '15 at 7:48

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