Is it?
It seems to be closed and there does seem to be inverses in it...?
Like
No.
For one thing you have to realize that the $\overline{1}$ in $\Bbb Z/2\Bbb Z$ isn't the "same object" as the $\overline{1}$ in $\Bbb Z/3\Bbb Z$. I prefer the notation $[k]_n$ to indicate the modulus when speaking of distinct quotients of $\Bbb Z$.
By definition:
$[1]_2 = \{\dots,-5,-3,-1,1,3,5,7,\dots\} = 1 + 2\Bbb Z$, while:
$[1]_3 = \{\dots,-5,-2,1,4,7,\dots\} = 1 + 3\Bbb Z$.
As you can see 4 is in the 2nd coset, but not the first, so these aren't even the same subset of the integers.
Secondly, the GROUP OPERATION in the two sets is completely different (even though we use the same symbol for both).
In $\Bbb Z/2\Bbb Z$, we have:
$\overline{k} + \overline{m} = \overline{k+m} \stackrel{\text{def}}{=} k+m\text{ (mod }2)$
In $\Bbb Z/3\Bbb Z$, we have:
$\overline{k} + \overline{m} = \overline{k+m} \stackrel{\text{def}}{=} k+m\text{ (mod }3)$.
Another way to see that $\overline{1}$ is different in the two groups, is that it has order 2 in the first group, but order 3 in the second group. It would be a good idea to convince yourself that if $g \in G$ has order $k$, and $H$ is a subgroup of $G$ with $g \in H$ as well, that $g$ STILL has order $k$ in $H$.
Finally, Lagrange's theorem states that ANY subgroup of a finite group has order dividing the order of the entire group. As 2 does not divide 3, no group of order 3 can have a subgroup of order 2.
(And the subset $\{\overline{0},\overline{1}\}$ is NOT closed in $\Bbb Z/3\Bbb Z$: in $\Bbb Z/3\Bbb Z$ we have:
$\overline{1} + \overline{1} = \overline{2}$ since 2 is congruent to 2 (mod 3), and NOT congruent to (0 mod 3) - which would mean, by definition, that 2 - 0 = 2 is a multiple of 3).
Wow thanks a lot!
That's many ways to answer my question, and really helps. I've been using modulus for a while now and I only just realised they are quotient groups haha
I've never heard of the modulo being represented as cosets but that's actually a good way to look at it. It makes so much more sense now...
Our lecturer only said to think of the (mod) as a clock and the numbers just go back to the start after you reach the modded number, and then moves on to teach quotient groups but never made the connection.
I'll see if I can prove that if then g has the same order in H and G. If not I'll probably make another thread