I have this question:
Let $G$ be a group, $a,b\in G$ and let $H$ be a subgroup of $G$.
i) Give the definition of the coset $aH$
ii) Prove that $aH = bH$ if and only if $a^{-1}b\in H$
iii) Suppose $H = \{1,h\}$ has exactly $2$ elements. If $H \unlhd G$ prove that $gh=hg$ for every $g\in G$
My attempts: i) $aH = \{ah| h \in H\}$
ii) Let me just roll with it, this is sort of what I was getting at(but didn't know how to do it):
($\to$) $aH = bH$, $ah = bh$, so $h = a^{-1}bh$, since obviously $h\in H$, $a^{-1}bh \in H$, but we also not $h=a^{-1}bh=h \to 1=a^{-1}b$, so $a=b$
($\leftarrow$) $a^{-1}b\in H$ then $a^{-1}b = h$, $b=ah$, $bH=aH$, that last step is probably not valid at all. Lost there
iii) $H \unlhd G$, so $g^{-1}hg\in H$
We have two values of $h$
$h=1$ $g^{-1}hg=g^{-1}*1*g=g^{-1}g=1$, obviously here $g*1=1*g$, $g=g$
$h=h$ $gh=hg$ is trivial case(Actually it might not be since it isn't abelian).
So what if it doesn't commute: $g^{-1}hg=h$, $hg=gh$ hence they commute.
Is any of this actually right?
The definition of $aH$ is correct.
For (ii), suppose first that $aH = bH$. Since $1\in H$, therefore, $b = b1\in bH$ and, since $aH=bH$, it follows that $b\in aH$. Thus, there is an element $h\in H$ such that $b = ah$. Multiplying by $a^{-1}$ on the left, we get $a^{-1}b = a^{-1}ah = h\in H$.
Conversely, assume that $a^{-1}b\in H$. Then $b = 1b = (aa^{-1})b = a(a^{-1}b) = ah_{1} \in aH$, where $h_{1} = a^{-1}b$. Now take any $bh_{2}\in bH$. Then $bh_{2} = (ah_{1})h_{2} = a(h_{1}h_{2})\in aH$, since $h_{1}h_{2}\in H$. This shows that $bH\subseteq aH$. You can fill in a similar argument to show that $aH\subseteq bH$.
For (iii), the assumption that $H\unlhd G$ tells you that $g^{-1}Hg = H$, for any $g\in G$. This means (just writing it out) that $$\{g^{-1}1g, g^{-1}hg \} = \{ 1, h \},$$ for any $g\in G$. Now, suppose that $g\in G$ is arbitrary. You want to show that $hg = gh$. Can you simplify the left hand side of the displayed set equality to help you come to this conclusion?