Let $H$ be a subgroup of a group $G$ and $a,b \in G$. Prove that the following Let $H$ be a subgroup of a group $G$ and $a,b \in G$. Prove that the following statements are equivalent. 
(1) $a^{-1}b \in H$
(2) $b^{-1}a \in H$
(3) $aH=bH$
So, I started with (3). $aH=bH \to a^{-1}aH=a^{-1}bH \to H=a^{-1}bH \to a^{-1}b \in H$ which is (1)
Then from second step above: $a^{-1}aH=a^{-1}bH \to H=a^{-1}bH \to b^{-1}aH=b^{-1}aa^{-1}bH \to b^{-1}aH=H \to b^{-1}a \in H$ which is (2)
I guess my question is, does my logic seem correct?
 A: It appears to be a correct start. You've shown $(3) \implies (1), (2)$, make sure you show those are equivalent and that one of those implies $(3)$. 
Here's another way of looking at it.
(1) $\implies$ (2): Suppose $a^{-1}b \in H$, then $a^{-1}b = h$ for some $h \in H$ and $$a^{-1}b = h \Leftrightarrow b = ah \Leftrightarrow 1 = b^{-1}ah \Leftrightarrow h^{-1} = b^{-1}.a$$ Since $h^{-1} \in H$ by definition of subgroup, $b^{-1}a \in H$.
(2) $\implies$ (3): Suppose $b^{-1}a \in H$. then $b^{-1}a = h$ for some $h \in H$ (which tells us $a = bh$ and $b = ah^{-1}$). Let $x \in aH$, then $x = ah_1$ for some $h_1 \in H$, but this means $x = ah_1 = bh h_1 \in bH$ which tells us $aH \subseteq bH$. Similar reasoning shows $bH \subseteq aH$.
(3) $\implies$ (1): Suppose $aH = bH$, then since $a = a \cdot 1 \in aH = bH$, there exists $h \in H$ such that $a = bh$. Observe that $$a = bh \Leftrightarrow 1 = a^{-1}bh \Leftrightarrow h^{-1} = a^{-1}b$$ which tells us $a^{-1}b \in H$, as required.
A: It seems correct,but it might be easier to begin with (a) ---> (b) -----(c) ----> (a). However,now that you began with (c)  ---(a) --->(b), now finish it with (b) ---->(c) as follows: Assume  $b^{-1}a \in H$. Then $H = b^{-1}aH$ and it's a simple matter to apply b to both sides to obtain (c). 
We also have to show (a) --->(c) and (b) ---->(c) for the proof to be complete. Robert Cardona already gave the proofs of these in his response. 
It's important to realize when a theorem states that n statements are equivalent, you have to show that each statement implies all the others or the proof is not complete.   
A: It's good, assuming that you have already proved that $xH=yH$ implies $zxH=zyH$, for $x,y,z\in G$.
However, you just need to prove that (1) is equivalent to (3), because the equivalence between (2) and (3) follows by observing that $aH=bH$ is the same as $bH=aH$.
(1)$\implies$(3)
Suppose $x=a^{-1}b\in H$ and let $h\in H$. Then
$$
ah=axx^{-1}h=aa^{-1}bx^{-1}h=b(x^{-1}h)\in bH.
$$
Therefore $aH\subseteq bH$.
The reverse inclusion follows from $x^{-1}=b^{-1}a\in H$.
(3)$\implies$(1)
Suppose $aH=bH$. Then $b=ah$ for some $h\in H$ and so $a^{-1}b=h\in H$.
