Which elements in $\mathbb{Z}_5$ have multiplicative inverses?

Question

Which elements in $\mathbb{Z}_5$ have multiplicative inverses?

I have drawn this in a small table:

*|0 1 2 3 4
-----------
0|0 0 0 0 0
1|0 1 2 3 4
2|0 2 4 1 3
3|0 3 1 4 2
4|0 4 3 2 1

Everywhere in table where we see a $1$ we have multiplicative inverse I think, so in total we have $4$ multiplicative inverses, is that right?

Is there a more efficient way answering this question? Because making table is too exhausting.

Answer 1

Indeed, from reading your table, we can see that every element with a $1$ in its row has a multiplicative inverse, which is to say all the non-zero elements.

However: in any book that mentions the notion of a multiplicative inverse, you should find the following statement in one form or another

$a$ has a multiplicative inverse in $\Bbb Z_n$ if and only if $a$ is relatively prime to $n$

Of course, if $n$ is prime, then every non-zero element has a multiplicative inverse in $\Bbb Z_n$.

Answer 2

Because 5 is prime, gcd(a,5)=1 for a=1,2,3,4. So 4 elements have multiplicative inverse (which means that $\mathbb Z_5$ is a field.

Answer 3

I will prove that if $p$ is a prime number (as $p=5$ in your case), then every $a \pmod p$ different from $0 \pmod p$ (i.e. we can take $1≤a≤p-1$) has a multiplicative inverse.

Since $p$ is a prime number, the gcd of $a$ and $p$ is $1$. By Bézout's identity, there are integers $x,y \in \Bbb Z$ such that $ax+py=1$. Then $$\bar x \bar a = \bar 1,$$ where $\bar k$ denotes the equivalence class of $k$ modulo $p$.

So we've shown that $a \pmod p$ has inverse $x \pmod p$.

Answer 4

Hint $\ $ Ponder the following chain of equivalences mod $n$

$\qquad\ \ ab\equiv 1\ \ {\rm for\ some}\ b$
$\iff a\mapsto ax\ \ \rm is\ onto.\ \ \ Proof\!:\ (\Leftarrow)\ \ clear.\ \ (\Rightarrow)\ \ c \equiv a(bc). $
$\iff a\mapsto ax\ \ \rm is\,\ 1\!-\!1,\ \ $ since, by pigeonholing, a map on a finite set is onto $\!\iff\!1\!-\!1$
$\iff \ker(a\mapsto ax) = 0$
$\iff ax\equiv 0\,\Rightarrow\, x\equiv 0$
$\iff\ n\mid ax\,\Rightarrow\ n\mid x$
$\iff (n,a) = 1$