# Math Help - index

1. ## index

If H is a subgroup of K, K is a subgroup of G, then |G:H| = |G:K||K:H| (do not assume G is finite).

2. Originally Posted by dori1123
If H is a subgroup of K, K is a subgroup of G, then |G:H| = |G:K||K:H| (do not assume G is finite).
Let $\{ a_i K | 1\leq i\leq n\}$ be cosets of $K$ in $G$ and let $\{ b_j H | 1\leq j \leq n \}$ be cosets of $H$ in $K$.
Show that all the cosets of $H$ in $G$ are given by $\{ a_i b_j H\}$.

3. that's the hint given in the book, but I don't get it...

4. Originally Posted by dori1123
that's the hint given in the book, but I don't get it...
Let $aH$ be a left coset where $a\in G$. We know that $a \in a_i K$ for some $a_i \in G$ since $\{ a_iK\}$ is a complete collection of cosets. Therefore $a = a_i b$ for some $b\in K$. Now $b\in b_j H$ for some $b_j \in K$ since $\{ b_j H\}$ is a complete collection of cosets. Therefore, $b = b_j h$ for some $h\in H$. Thus, $a = a_i b = a_i b_j h \implies a \in a_i b_j H$. We have shown that $\{ a_i b_j H | 1\leq i \leq n , 1\leq j \leq m \}$ contains all cosets of $H$ in $G$. Therefore, there are at most $n\cdot m$ cosets of $H$ in $G$. The reason why at most and not exactly is because it is possible that different $a_i$'s and $b_j$'s can produce the same coset. It remains to show that if $a b H = a' b'H$ then $a = a',b=b'$. Can you finish the last step?

5. Originally Posted by ThePerfectHacker
It remains to show that if $a b H = a' b'H$ then $a = a',b=b'$. Can you finish the last step?
So if $abH = a'b'H$, then $abh_1 = a'b'h_2$ for some $h_1, h_2 \in H$, and so $abh_1h_2^{-1}(b')^{-1}(a')^{-1} = 1 \iff h_1h_2^{-1} = 1, b(b')^{-1} = 1,$ and $a(a')^{-1} = 1$. So $a = a'$ and $b = b'$.
Is this correct? does it make sense?

6. Originally Posted by dori1123
Is this correct? does it make sense?
This is how to do it. If $abH = a'b'H$ then it means $aK = a'K$ (why?) and therefore $a = a'$. From here it follows that $b=b'$.