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Let $G$ be a finite group. Let $H_2\subseteq H_1$ be subgroups.

Let $R$ be a complete system of representatives for the left cosets of $H_1$ in $G$.
Let $S$ be a complete system of representatives for the left cosets of $H_2$ in $H_1$.

Construct a complete system of representatives for the left cosets of $H_2$ in $G$.

I tried to do this myself below. I'm not sure if it is correct.

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1 Answer 1

As $G$ is finite, $G/H_1=\{g_1H_1,...,g_kH_1\}$ and $H_1/H_2=\{h_1H_2,...,h_lH_2\}$.
Which gives $R=\{g_1,...,g_k\}$ and $S=\{h_1,...,h_l\}$.
Let $T=\{g_α\cdot h_β :α∈ℕ_{≤k},β∈ℕ_{≤l}\}$.

I'm going to prove that $T$ is the set asked for.
Let $g∈G$, then $g∈g_α H_1$, then $∃h∈H_1:g=g_αh$.
Let $h∈H_1$, then $h∈h_β H_2$, then $∃h_2∈H_2 :h=h_βh_2$.
Then $g=g_ah_bh_2$, so $g∈g_αh_βH_2$.

Therefore the set $T$ contains for every $g∈G$ a representative for $g$ in the form of a left coset of $H_2$ of $G$.

I still need to prove that $T$ doesn't contain too much representatives (double representatives). In this prezi I proved that:
$[G:H_2]=[G:H_1][H_1:H_2]$ (this prezi is in dutch, but I think it will be clear for english readers)

So $|T|$ must equal $|S|\cdot |R|=k\cdot l$. But this follows directly from

$$T=\{g_α\cdot h_β :α∈ℕ_{≤k},β∈ℕ_{≤l}\}$$

QED

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