# Prove that these two integer groups have equivalent Cayley tables.

This is an extension of a question from M.A. Armstrong's "Groups and Symmetry" (1988).

7.1 Check that the numbers 1, 2, 4, 5, 7, 8 form a group under multiplication modulo 9 and show that this group is isomorphic to $\mathbb{Z}_6$.

First, I computed the Cayley table for the $\mathbb{Z}_9$ subgroup and determined that these six elements do in fact form a group. I did the same for $\mathbb{Z}_6$ both under multiplication and addition, and did not find a match for the subgroup mentioned above.

Fine. I figure that there are a number of permutations of Cayley tables for any given group, so I'll just play around with a few until the elements appear congruent i.e. taking the same place in the table. After doing this for 3 tables, I decided to change up my approach.

I remembered that the power of each element was an important part of the structure of the group, so I counted the number of steps it takes for each element to reach identity. For $\mathbb{Z}(6,+): 0 = 1^6 = 2^5 = 3^4 = 4^3 = 5^2$ where e.g. the 1 element takes six iterations to reach 0 under addition modulo 6.

After doing the same for the six element $\mathbb{Z}_9$ subgroup, I found that the elements had congruent powers. I thought this is a good sign. Here is what I found: $1 = 2^4 = 4^5 = 5^2 = 7^3 = 8^6$ where again the power represents the number of operations taken to reach identity, in this case under multiplication modulo 9.

Equating the elements of the same power takes the first map of $\mathbb{Z}_6$ to the $\mathbb{Z}_9$ subgroup thus: $0 \rightarrow 1; 1 \rightarrow 8; 2 \rightarrow 4; 3 \rightarrow 2; 4 \rightarrow 7; 5 \rightarrow 5$.

My question is if the powers of the elements correspond, why don't they form equivalent Cayley maps. The patterns of each are totally different!

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The powers are wrong in $\mathbb Z(6,+)$. It goes like this; 0 = 1^6 = 2^3 = 3^2 = 4^3 = 5^6. Also for $\mathbb Z_9$ they should be $1=2^6=4^3=5^6=7^3=8^2$. You must be doing something wrong. –  wircho Jan 6 '12 at 16:31
You also did some mistakes in calculating the orders in $\mathbb{Z}_9$. Note that $$8 = -1 \mod 9$$ so it has an order of 2... Also $2^4=16 \neq 1 \mod 9 \,$ and $5^2 \neq 1 \mod 9$....$4^5=2^{10}=1024= 7 \mod 9$... And last $7^3=(-2)^3=-8=1$ so this is the only right one.... –  N. S. Jan 6 '12 at 16:33
Sorry for repeating what you said N.S. I edited my comment without seeing yours. –  wircho Jan 6 '12 at 16:36

The groups are in fact equivalent, because there exists an isomorphism between these groups. For finite groups, a nice way to explain the notion of isomorphism is that there is a way to write both Cayley tables and associate one member of one table to the member of the other table, in a way that the tables coincide. I'll explain what I mean with your example. $\mathbb Z_6 = \{ 0, 1, 2, 3, 4, 5 \}$ with addition, and your subgroup of $\mathbb Z_9$ with multiplication can be ordered in the following manner : $H = \{ 1, 2, 4, 8, 7, 5 \}$. Now consider $\phi : \mathbb Z_6 \to H$ which maps $$\begin{pmatrix} 0 \\ 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ \end{pmatrix} \mapsto \begin{pmatrix} 2^0 \\ 2^1 \\ 2^2 \\ 2^3 \\ 2^4 \\ 2^5 \\ \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 4 \\ 8 \\ 7 \\ 5 \\ \end{pmatrix}.$$ (Here, read things modulo $9$, i.e. $2^5 = 5$ when this equation is read modulo $9$.) Now it is clear that they have the same multiplication table since when $i + j = k$, $2^i \times 2^j = 2^{i+j} = 2^k$. If you write out the Cayley table and rename the integers $1$, $2$, $\dots$, $7$ and $5$ by $2^0$ , $\dots$, $2^5$ $(\mathrm{mod} \ 9)$, you'll see that the Cayley tables are indeed the same. In this context $\phi$ is called an isomorphism and the two groups are called isomorphic. $$\begin{bmatrix} + & 0 & 1 & 2 & 3 & 4 & 5 \\ 0 & 0 & 1 & 2 & 3 & 4 & 5 \\ 1 & 1 & 2 & 3 & 4 & 5 & 0 \\ 2 & 2 & 3 & 4 & 5 & 0 & 1 \\ 3 & 3 & 4 & 5 & 0 & 1 & 2 \\ 4 & 4 & 5 & 0 & 1 & 2 & 3 \\ 5 & 5 & 0 & 1 & 2 & 3 & 4 \\ \end{bmatrix} \sim \begin{bmatrix} \times & 2^0 & 2^1 & 2^2 & 2^3 & 2^4 & 2^5 \\ 2^0 & 2^0 & 2^1 & 2^2 & 2^3 & 2^4 & 2^5 \\ 2^1 & 2^1 & 2^2 & 2^3 & 2^4 & 2^5 & 2^0 \\ 2^2 & 2^2 & 2^3 & 2^4 & 2^5 & 2^0 & 2^1 \\ 2^3 & 2^3 & 2^4 & 2^5 & 2^0 & 2^1 & 2^2 \\ 2^4 & 2^4 & 2^5 & 2^0 & 2^1 & 2^2 & 2^3 \\ 2^5 & 2^5 & 2^0 & 2^1 & 2^2 & 2^3 & 2^4 \\ \end{bmatrix}$$ Hope that helps,
This is just because the group $H$ is cyclic, so matching Cayley tables simplies to matching generators. $2$ was the generator in this case. –  Patrick Da Silva Jan 6 '12 at 17:12
Indeed =) For a cyclic group, sending a generator to another generator is always an automorphism (i.e. an isomorphism from $G$ to itself). In another words, the same trick works if you replace all the two's in the table on the left by $5$'s and rename the elemnts $1,2,4,5,7,8$ by $5^0 = 1, 5^1 = 5, 5^2 = 7, 5^3 = 8, 5^4 = 4, 5^5 = 2$. –  Patrick Da Silva Jan 6 '12 at 18:00
since the groups are cyclic, you just need to find a generator for each. for $\mathbb{Z}/(6)$ is generated by 1. now find one of 1,2,4,5,7,8 which has order 6 under multiplication (such as 2: 2,4,8,16=7,32=5,64=1). if the only idea of isomorphism you have is matching multiplication tables, then you can write the tables equating 1 mod 6 with 2 mod 9 and completing.