Commutativity among the elements in the quotient group $G$ and the ful group $H$ under homomorphism $H \overset{r}{\to} G$ Say the full group is $H$, and we pick up a normal subgroup $N$, and we define the quotient group $$G=H/N.$$
There is a group homomorphism $r$ from $H$ to $G$
$$
H \overset{r}{\to} G.
$$
My question concerns the following relations are true in general or whether there are counter examples?


*

*$ g \cdot r(h) \cdot g^{-1}=r(h)$,  for $\forall h \in H$, $\forall g \in G$.

*$ r(h) \cdot g \cdot r(h)^{-1}=g$, for $\forall h \in H$, $\forall g \in G$.
If some of them is not true, please give the simplest (counter) example (especially the finite group).
 A: Since $r$ is a group homomorphism, your relations are in general not true: Try a pair $(h_1,h_2)$ of (non-commuting elements), of a non-abelian group $H$, such that: $h_1h_2h_1^{-1}h_2^{-1}\notin N$,  for providing counterexamples: 
$$
r(h_1h_2)=r(h_1)r(h_2)=r(h_2)r(h_1)=r(h_2h_1)\Leftrightarrow \\r(h_1)r(h_2)r(h_1)^{-1}r(h_2)^{-1}=1_N
\Leftrightarrow \\ 
r(h_1h_2h_1^{-1}h_2^{-1})=1_N=r(N)
\Leftrightarrow \\ 
h_1h_2h_1^{-1}h_2^{-1}\in N
$$
thus, (taking the contrapositive of the above) we get:
$$
h_1h_2h_1^{-1}h_2^{-1}\notin N \Leftrightarrow r(h_1)r(h_2)=r(h_1h_2)\neq r(h_2h_1)=r(h_2)r(h_1)
$$
Maybe the simplest counterexample (reflecting the above argument in its most simplistic  form) is one already mentioned by user  Jason DeVito in his comment above: Pick any non-abelian group $H$ and let $N=\{e\}$. It is easy to see that, in that case: $$H= H/N= H/\{e\}$$
Thus, any pair of non-commuting elements of $H$ would do. 
A: They do not hold in general.
Consider $H= D_8 = <\sigma, \tau | \sigma^8 = 1, \tau^2 = 1, \tau \sigma^i \tau = \sigma^{8-i}>$ and consider its normal subgroup $N=<1, \sigma^4>$.
Now $G=H/N$ has order $8$ and has elements $1, \tau, \sigma, \sigma^2, \sigma^3, \sigma\tau, \sigma^2 \tau,\sigma^3 \tau$. 
Consider $r:H \to G$ the usual projection. Now $g=\tau, h=\sigma^5\tau$ gives you
$$\tau r(\sigma^5\tau) \tau = r(\tau\sigma^5) = \tau\sigma$$
while $r(\sigma^5 \tau) = \sigma\tau$. You can do the same for the other relation.
