1. Jul 5, 2013

### Someone2841

Hello. I have been looking into group theory for its applications to subject I am studying. I am not a mathematician by profession or training, but I find it has great use to any analytical pursuit. With that said, I have outlined below type of group that I would like to know more about. For example: what other groups are like it?, is there anything strange about the way I've defined this group?, what would help in my understanding of this group and groups like it?, etc. Thanks in advance!

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Let $C$ be the cyclic ring of integers mod 12. Please note that arithmetic with elements of $C$ and its subsets will be assumed to be mod 12 unless otherwise indicated. Define the set $M_{k \in C}$ as
$\{x : n \in \mathbb{Z}, k-1 \leq n \leq k+5, x \equiv 7n\}$​

For example, $M_0 = \{0,2,4,5,7,9,11\}$. These subsets $M_k$ clearly do not form subgroups of $C$; however, a group structure can be imposed on them. Define the action $S_k$ on $M_k$ as

$S_k:M_k×M_k,m≡7(x+k−1)↦7((x+_7 1)+k−1)$**​

Where $+_7$ is addition mod 7. For example, $S_0$ would map 4 onto 11 since $x=3$ in the equation $4 \equiv 7(x + k - 1)$ and $7((x +_7 1) + k - 1) \equiv 11$. Two iterations of the action (i.e., $S^2_k$) would then map each element to the next element (e.g., $9 \mapsto 11, 11 \mapsto 0, 0 \mapsto 2,$ etc. in $M_0$).

The set $\{S^n_k\}_{n \in Z, 0 \leq n \leq 6}$ forms a cyclic action group (with $S_k$ as the group generator) on $M_k$. The identity action is $S^0_k \doteq S^7_k$ and the inverse action is $S^{-n \mod{7}}_k$.

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**Edited from x-k+1 to x+k-1

Last edited: Jul 5, 2013
2. Jul 5, 2013

### micromass

May I ask what $11$ gets mapped to under $S_0$?

3. Jul 5, 2013

### Someone2841

I made a mistake. $S_k$ should be defined as

$S_k:M_k×M_k,m≡7(x+k-1)↦7((x+_7 1)+k-1)$​

in order to deal with the very case of $S_0(11)$. In the equation $11 \equiv 7(x + k - 1)$, $x = 6$ and so $7((x+_7 1)+k-1) \equiv 5$. With this correction, $11$ gets mapped to $5$ under $S_0$. Sorry for the mistake!

4. Jul 9, 2013

### Stephen Tashi

The only context that I'm familiar with for "actions" on sets is that there is a definition for a group action on a set. (Google tells me that there is also definition of a ring action - and a neat definition of a module: http://mathoverflow.net/questions/100565/why-are-ring-actions-much-harder-to-find-than-group-actions. ) To define a group action, you'd normally start with the group and show how it acts on a set. I think what you should do is define the "functions" $S_k$ and prove the set of functions forms a group. The abstract group that it forms then has the "group action" on the set specified by the functions.

I don't know what part of your result depends only on having a ring-with-identity, a commutative ring with identity, or a field. You might think about whether a similar construction works when applied to mod 6 arithmetic.