Question 1. Where's the subgroup H in the left picture?

It's not there. If it was, it would be a circle containing the identity $e$.

Question 2. Is my drawing right for gH? What's the circle around g?

Yes, it looks correct. The circle around $g$ is the coset $gH$.

Question 3. How do you see what page 143 says about 'all the $g^{−1}$ arrows lead back from the left coset gH to the subgroup H'?

The best way to think about cosets of a subgroup $H$ are as partitions of the group $G$ into equally sized pieces of size $|H|$. These pieces are permuted when multiplied on the right (or left) by group elements, with the condition that multiplying by an element of $H$ is the trivial permutation. So the $g_1$ arrows simply represent how $H$ has been shifted by $g_1$ multiplication.

Then when we form the *factor group*, this simply means we form a new group consisting of cosets (i.e. partition pieces) with group multiplication given by the permutations.

Consider the map $\sigma\colon G\to G$ defined by $\sigma(x)=x^2$. This is a homomorphism, because $G$ is abelian, being cyclic. The kernel of $\sigma$ is
$$
\ker\sigma=\{x\in G:x^2=1\}
$$
while its image is $H$.

If the order of $G$ is odd, no element $x\ne1$ can have the property that $x^2=1$, so $\sigma$ is injective, hence surjective. So, in this case, $H=G$ and, of course, there's only one coset.

If the order $n$ of $G$ is even, there is *exactly* one element $x\in G$ such that $x^2=1$ and $x\ne 1$, precisely $a^{n/2}$. (Why?) So the kernel of $\sigma$ has two elements and $|H|=|G|/|\ker\sigma|=n/2$. Therefore $|G|/|H|=2$ and so there are exactly two cosets; since $a\notin H$ (why?), the cosets are
$$
H\quad\text{and}\quad Ha.
$$

If $G$ is infinite, then $G\cong\mathbb{Z}$ and the image of $H$ under this homomorphism is $2\mathbb{Z}$. Again the cosets are
$$
H\quad\text{and}\quad Ha.
$$

## Best Answer

Lets work with smaller symmetric groups. Let's let $G=S_3$, $H=S_2$. We can actually write out each group in set notation;

$$S_3=\{(1), (12), (13), (23), (123), (132)\}$$

$$S_2=\{(1), (12)\}$$

Now, for any $g\in S_3$, a left coset is a set $gH$. What does that mean? Well, lets look at $g=(23)$. We can "multiply" all elements in $S_2$ by $(23)$. Of course, when working with the symmetric group, we are composing, not multiplying, but you get the idea i hope. Then

$$(23)H=(23) \{(1),(12)\}=\{(23)(1), (23)(12)\}=\{(23),(132)\}$$

This is one of the left cosets of $H=S_2$ in $S_3$. The same principle can be applied to your problem, but of course there are a lot more elements in $S_6$ and $S_7$, so now using @C Monsour's answer should help to enlighten a bit more...