Is it surjective because of $K \subset H$?

Question

Let $G$ be a group and $H,K$ normal subgroups of $G$ s.t. $K \subset H$. If I define the following map: $\phi:G /K \to G/H, gK \mapsto gH$ then the claim is that it is surjective. Is it because we have that $K \subset H$ and hence $G /H \subset G/K$?

Answer 1

The fact that the application is surjective is trivial, since taken $gH\in G/H$, then $\phi(gK)=gH$. The tricky part is proving that this is actually a well posed application. And this can be done as follows: $$aK=bK\Rightarrow ab^{-1}\in K\subset H \Rightarrow ab^{-1}\in H \Rightarrow aH=bH \Rightarrow \phi(aK)=\phi(bK)$$ Now you have just to prove that is an homomorphism (trivial), find its kernel and then apply first isomorphism theorem.

Answer 2

More precisely, this is because of the $3^{ rd }$ isomorphism theorem: $H/K$ is a normal subgroup of $G/K$, and $$(G/K)\big/(H/K)\simeq G/H.$$

What can be said of $G/H$ as compared to $G/K$ is that every $H$-coset is a union of $K$-cosets.

Answer 3

Since we aim to get a map between coset sets, but we can't avoid using coset representatives, we need to ensure that $g'\in gK\Rightarrow \phi(g'K)=\phi(gK)$ (good definition). But this is indeed the case, because by $K<H$ (see $(*)$ step):

\begin{alignat}{1} g'\in gK &\Rightarrow g^{-1}g'\in K \\ &\stackrel{(*)}{\Rightarrow} g^{-1}g'\in H \\ &\Rightarrow g'\in gH \\ &\Rightarrow g'H=gH \\ &\Rightarrow \phi(g'K)=\phi(gK) \\ \end{alignat}

About surjectivity, notice that, by $K<H$, it is $\#G/K=[G:K]=[G:H][H:K]>[G:H]=\#G/H$ and use the pigeonhole principle.

Finally, as long as you interpret $G/H$ and $G/K$ as quotient sets (not necessarily groups), the normality of $H$ and/or $K$ is not necessary for the map to be (well defined and) surjective.