# Sum of all invertible integers modulo a prime

If $p$ is an odd prime, I want to show that $$\sum_{\beta \in \mathbb{Z}_p^\ast} \beta^{-1} = \sum_{\beta \in \mathbb{Z}_p^\ast} \beta = 0$$

Now I know $\mathbb{Z}_p^\ast = \{1,\dots,p-1\}$, so $$\sum_{\beta \in \mathbb{Z}_p^\ast} \beta = \sum_{i=1}^{p-1}i = \frac{(p-1)p}{2}.$$ Since $p$ is odd, $2 \mid p-1$ and so this is the integer $pq$, where $q = (p-1)/2$. As $p \mid pq$, it follows that $pq \equiv 0 \mod p$ so I have that $$\sum_{\beta \in \mathbb{Z}_p^\ast} \beta = 0.$$

I know that the set of invertible elements modulo a prime is equivalent to the set of their inverses so $$\sum_{\beta \in \mathbb{Z}_p^\ast} \beta^{-1} = \sum_{\beta \in \mathbb{Z}_p^\ast} \beta$$ only differs by the order in which the elements of $\mathbb{Z}_p^\ast$ are added, but I am not sure how to explicitly show this.

Would a viable approach be to show that there is a bijective map $f: \mathbb{Z}_p^\ast \rightarrow \mathbb{Z}_p^\ast$ such that $\beta \mapsto \beta^{-1}$?

• You are doing things very well. Bijective map is fine, every $a$ in the interval $[1,p-1]$ has a unique inverse modulo $p$. You probably should not quote the sum of first $p-1$ integers so freely, need to make sure it makes sense in the algebraic setting. Aug 19, 2013 at 2:46
• It would be cleaner to pair every $k$ with its additive inverse. Aug 19, 2013 at 2:53
• When you say that I need to make sure it makes sense in the algebraic setting, do you mean that the sum of the elements in $\mathbb{Z}_p^\ast$ is equivalent to the sum from $1$ to $p-1$ or that the sum from $1$ to $p-1$ is $p(p-1)/2$?
– Alex
Aug 19, 2013 at 2:55
• The latter, particularly the division by $2$. Aug 19, 2013 at 2:57
• If we are working in a very number-theoretic style, with congruences, there is not much problem. But in a more algebraic style, one has to worry about how the real deivision by $2$ us connected to $2^{-1}$. I seriously suggest you use the pairing argumentm ny second comment. Much more algebraic. Aug 19, 2013 at 3:04

If you can show the inverse obeys $(x^{-1})^{-1} = f^{2}(x) = f(f(x)) = x$, then $f$ is a bijection because $f^2$ is.

• For this I would just need to show that $f^2$ is the identity map?
– Alex
Aug 19, 2013 at 3:09
• Yes, and the identity map is clearly a bijection. Any map that isn't a bijection would have a smaller number of elements in its in image, so $f$ is a bijection too. Aug 19, 2013 at 3:12