# How do generators of a group work?

$$G$$ is a group, $$H$$ is a subgroup of $$G$$, and $$[G:H]$$ stands for the index of $$H$$ in $$G$$ in the following example:

Let $$G=S_3$$, $$H=\left<(1,2)\right>$$. Then $$[G:H]=3$$.

I know the definition of group generators: A set of generators $$(g_1,...,g_n)$$ is a set of group elements such that possibly repeated application of the generators on themselves and each other is capable of producing all the elements in the group.

What does the individual elements of $$H=\left<(1,2)\right>$$ look like? Any help would be greatly aprciated.

P.S. I know how to find the index when the groups don’t involve a group generator, the thing I need help with is understanding the group generator.

• What's the actual question here? – Lord Shark the Unknown Dec 10 '18 at 4:18
• My question is more or less this: What does the individual elements of $H$ look like? – AMN52 Dec 10 '18 at 4:21
• $H$ must have the identity element. It must also have the element $(1\ 2)$. – Lord Shark the Unknown Dec 10 '18 at 4:22
• $(1,2)$ has order 2 so it's the only non identity element in $H$. – Justin Stevenson Dec 10 '18 at 4:22
• Why does $H$ only contain $(1,2)$ and the identity element? – AMN52 Dec 10 '18 at 4:25

$$(12)$$ is a transposition. $$(12)^{-1}=(12)$$, that is, it's its own inverse. All that can be gotten by taking powers of $$(12)$$ is $$(12)$$ and $$e$$, the identity. (Note: In general, $$\langle a\rangle =\{a^n:n\in\mathbb Z\}$$). Thus $$\langle (12)\rangle =\{(12),e\}$$, a two element group.
Since $$\mid S_3\mid=6$$, we get $$[S_3:\langle (12)\rangle] =3$$.
Because $$H$$ is the set of all elements of the form $$(1,2)^n$$ and $$(1,2)^2=e$$.