Proving the following conditions are equivalent I have to prove four conditions are equivalent. I'm guessing I should proceed with (a) implies (b), (b) implies (c), (c) implies (d), and (d) implies (a)?
I have gotten (a) implies (b), (b) implies (a), and (b) implies (d). I'm struggling with (b) implies (c) and (c) implies (d) and (d) implies (a) though.
Here is the problem and my work so far:
Let $ f: G → H $ be a homomorphism of groups, and let a and b be elements of G. Let K be the kernel of $f$. Prove that the following conditions are equivalent:
(a) $f(a)=f(b)$
(b) $a^{-1}b$ is in $K$
(c) $b$ is in the coset $aK$
(d) The cosets $bK$ and $aK$ are equal
I have so far: Suppose that $f(a)=f(b)$. Then $f(a^{-1}b) = f(a^{-1})f(b) = f(a)^{-1}f(b)=1.$ Therefore, $a^{-1}b$ is in the kernel K. If $a^{-1}b$ is in $K$, then $1 = f(a^{-1}b)=f(a)^{-1}f(b)$, so $f(a)=f(b)$. This shows that the first two conditions are equivalent. I have a lengthy argument that I won't put here for how (b) implies (d). 

Now how do I show that (b) implies (c), (c) implies (d), and (d) implies (a)?

Thanks.
 A: $\bbox[5px,border:2px solid #4B0082]{(b)\implies (c)}$
$$\begin{align} a^{-1}b\in K&\implies (\exists k\in K)(a^{-1}b=k)\\
&\implies (\exists k\in K)(b=ak)\\
&\implies b\in aK \end{align}$$
$\bbox[5px,border:2px solid #4B0082]{(c)\implies (d)}$
Suppose $b\in aK$. There exists $k\in K$ such that $b=ak$.


*

*$\bbox[5px,border:2px solid #7FFF00]{bK\subseteq aK}$
Let $x\in bK$. There exists $\overline k\in K$ such that $x=b\overline k$. It follows that $x=b\overline k=a\underbrace{k\overline k}_{\large \in K}$ and therefore $x\in aK$.

*$\bbox[5px,border:2px solid #7FFF00]{aK\subseteq bK}$
Let $x\in aK$. There exists $\overset{\sim}k\in K$ such that $x=a\overset{\sim}k$. Therefore $x=a\overset\sim k=b\,\underbrace{k^{-1}\overset{\sim} k}_{\large \in K}\in bK$.


It follows that $aK=bK$ as wanted.
$\bbox[5px,border:2px solid #4B0082]{(d)\implies (a)}$
$$\begin{align} aK=bK&\implies (\forall k\in K)(\exists k'\in K)(ak=bk')\\
&\implies (\forall k\in K)(\exists k'\in K)(a=bk'k^{-1})\\
&\implies (\forall k\in K)(\exists k'\in K)(f(a)=f(bk'k^{-1}))\\
&\implies (\forall k\in K)(\exists k'\in K)\left(f(a)=f(b)f(\underbrace{k'k^{-1}}_{\large \in \ker (f)})\right)\\
&\implies (\forall k\in K)(\exists k'\in K)(f(a)=f(b))\\
&\implies f(a)=f(b)\end{align}$$
