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No, replacing the representative element g with another element g' from the same coset does not yield a different coset. If g' belongs to the same coset as g (i.e., $g' \in gH$), then the coset generated by g' is identical to the coset generated by g ($g'H = gH$).

Conditions: g and g' are both elements of the group G; g' and g belong to the same coset

No, replacing the representative element g with another element g' from the same coset does not yield a different coset. If g' belongs to the same coset as g (i.e., $g' \in gH$), then the coset generated by g' is identical to the coset generated by g ($g'H = gH$).

Conditions: g and g' are both elements of the group G; g' and g belong to the same coset

A normal subgroup satisfies $gH = Hg$ because its defining property is that it is invariant under conjugation, meaning $gHg^{-1} = H$. By right-multiplying both sides of this equation by $g$, the $g^{-1}$ and $g$ cancel out, yielding the equality of the left coset $gH$ and the right coset $Hg$.

Conditions: H is a normal subgroup of G; g is an element of the group G

A normal subgroup satisfies $gH = Hg$ because its defining property is that it is invariant under conjugation, meaning $gHg^{-1} = H$. By right-multiplying both sides of this equation by $g$, the $g^{-1}$ and $g$ cancel out, yielding the equality of the left coset $gH$ and the right coset $Hg$.

Conditions: H is a normal subgroup of G; g is an element of the group G

We can intuitively understand a normal subgroup by visualizing the group G as an outer rectangle and the subgroup H as an inner rectangle. The condition $gHg^{-1} = H$ means that if you 'transport' the entire region H using an element g and then bring it back using $g^{-1}$, the resulting region still lies completely within H.

Conditions: H is a subgroup of G; g is an element of G outside H

We can intuitively understand a normal subgroup by visualizing the group G as an outer rectangle and the subgroup H as an inner rectangle. The condition $gHg^{-1} = H$ means that if you 'transport' the entire region H using an element g and then bring it back using $g^{-1}$, the resulting region still lies completely within H.

Conditions: H is a subgroup of G; g is an element of G outside H