Generating subgroups of $S_3$ I am trying to understand how the generation of subgroups work:
So if I have $\sigma_1= (123)$ and $\sigma_2=(12)$, both of these generate subgroups of $s_3$, how?
Do I give it an element, say $1$ and then see what elements occur with repeated iteration of $\sigma_i$ for subgroup $i$?

E.g. $1\sigma_1=2,1\sigma_1^2=3,1\sigma_1^3=1$ hence we generate $\{1,2,3\}$. Or do we put in $(123)$ and see what we obtain: E.g. $\{(231),(312),(123)\}$?

 A: The idea is that a group of must be closed and the inverse element of it must be in it. By the group "generated" by some elements we mean the smallest such subgroup which contains all the elements first considered. 
You are right. $(123) = \{ (123), (132), I \}$ does not generate $S_3$. 
But the two elements $ (123), (12) $ does. 
Let $G$ be the smallest group which has both $(123)$ and $(12)$ i.e. the subgroup generated by $A = \{  (123), (12) \}$ 
Notice that if $(123) (12) = (13)$ hence $(13) \in G$. Now since $G$ is a group it must also contain $(123) (13) = (23)$. 
Hence $G$ contains all elements of $S_3$ hence $ A  $ generates $G$. 

Now for any $a$ the set $\{a, a^2, a^3 \}$ forms a group given that $ a^3 = e $ and $ a^{m} \cdot a^{n} = a^{m + n}$. This is the generic cyclic group of order $3$. Try proving that $(a) =  \{a^{i} \ | \ i \in \Bbb Z\} $ is a group given the opration specified above. Now prove using the theory of division that if the order of $a$ is $3$ then $ (a) = \{a, a^2, a^3 \} $. 
What you have posted in your answer is the same thing with $a = (123)$
A: What you are dealing with is $\langle (123) \rangle = \{(\cdots,(123)^{-2},(123)^{-1},(123)^0=e,(123)^1,(123)^2,\cdots\}$
Which of course has elements repeating, because $(123)$ is of order $3$, and hence we obtain the elements $\{(123),(132),(1)(2)(3))\}=\langle(123)\rangle$.

Why is this a subgroup? It has identity, $(1)(2)(3)$, that does nothing.
