# Multiplication table for Z_n

1. May 8, 2010

### Combinatus

1. The problem statement, all variables and given/known data

Create a multiplication table for the group of invertible elements in the ring $$Z_{10}$$. Can you rename the elements and arrange them so that the multiplication table is transformed into a multiplication table for the group $$Z_n$$ for some n?

2. Relevant equations

3. The attempt at a solution

If $$p \in Z_m$$, $$p$$ has an inverse iff $$GCD(p,m)=1$$, so the invertible elements in $$Z_{10}$$ are 1, 3, 7 and 9, and we end up with

$$\begin{bmatrix} 1 & 3 & 7 & 9\\ 3 & 9 & 1 & 7\\ 7 & 1 & 9 & 3\\ 9 & 7 & 3 & 1\\ \end{bmatrix}$$

as the suspiciously matrix-looking multiplication table in $$Z_{10}$$.

I don't know what the second sentence of the problem implies though. After attempting proof by asking IRC, I received the reply "Z/10Z =~ Z/2Z x Z/5Z -> (Z/10Z)* =~ (Z/2Z)* x (Z/5Z)* =~ Z/4Z", but I haven't seen any similar notation before. Where do I begin on this, or perhaps, what should I read to get a better understanding of similar problems?

Last edited: May 8, 2010
2. May 8, 2010

### tiny-tim

Hi Combinatus!

(nice LaTeX, btw! )
Well, it's a 4x4 matrix, so it obviously can only be the table for Z5

so re-name 3 7 and 9 as some permutation of 2 3 and 4 so that the table works.

3. May 8, 2010

### Combinatus

Okay, so in $$Z_5$$ we get

$$\begin{bmatrix} 1 & 2 & 3 & 4\\ 2 & 4 & 1 & 3\\ 3 & 1 & 4 & 2\\ 4 & 3 & 2 & 1\\ \end{bmatrix}$$

The permutation that generates this table from the one of the inverses in $$Z_{10}$$ (i.e. the one in my previous post) is thus 1 -> 1, 3 -> 2, 7 -> 3, 9 -> 4.

The key to the problem, however, says that "If the elements are arranged to 1, 3, 9, 7 and they are named 1 -> 0, 3 -> 1, 7 -> 3, 9 -> 2, you get the table for the group $$Z_4$$." Even if you include 0 in the multiplication table for the group $$Z_4$$, they don't look similar at all, so I'm not sure what they're on about.

4. May 8, 2010

### tiny-tim

Hi Combinatus!

hmm … some of Z4 is right, but if 1 -> 0, then there should be 1s (for 0s) all along the top and left of the table (and there aren't)

the key is wrong … it must be Z5

5. May 8, 2010

### jbunniii

The notation in the key is a bit vague, but what it is saying is that $Z_{10}^*$ is isomorphic to $Z_5^*$ (the MULTIPLICATIVE group of units of $Z_5$), which is isomorphic to $Z_4$ (the ADDITIVE group). This is indeed true.