Commutativity among the elements in the quotient group $G$ and the ful group $H$ under homomorphism $H \overset{r}{\to} G$

Question

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?

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

2. $ 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).

Answer 1

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.

Answer 2

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.