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).

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- Oct 10th 2008, 09:23 AMdori1123index
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).

- Oct 10th 2008, 09:57 AMThePerfectHacker
Let $\displaystyle \{ a_i K | 1\leq i\leq n\}$ be cosets of $\displaystyle K$ in $\displaystyle G$ and let $\displaystyle \{ b_j H | 1\leq j \leq n \}$ be cosets of $\displaystyle H$ in $\displaystyle K$.

Show that all the cosets of $\displaystyle H$ in $\displaystyle G$ are given by $\displaystyle \{ a_i b_j H\}$. - Oct 10th 2008, 11:48 AMdori1123
that's the hint given in the book, but I don't get it...

- Oct 10th 2008, 12:18 PMThePerfectHacker
Let $\displaystyle aH$ be a left coset where $\displaystyle a\in G$. We know that $\displaystyle a \in a_i K$ for some $\displaystyle a_i \in G$ since $\displaystyle \{ a_iK\}$ is a complete collection of cosets. Therefore $\displaystyle a = a_i b$ for some $\displaystyle b\in K$. Now $\displaystyle b\in b_j H$ for some $\displaystyle b_j \in K$ since $\displaystyle \{ b_j H\}$ is a complete collection of cosets. Therefore, $\displaystyle b = b_j h$ for some $\displaystyle h\in H$. Thus, $\displaystyle a = a_i b = a_i b_j h \implies a \in a_i b_j H$. We have shown that $\displaystyle \{ a_i b_j H | 1\leq i \leq n , 1\leq j \leq m \}$ contains all cosets of $\displaystyle H$ in $\displaystyle G$. Therefore, there are

__at most__$\displaystyle n\cdot m$ cosets of $\displaystyle H$ in $\displaystyle G$. The reason why at most and not exactly is because it is possible that different $\displaystyle a_i$'s and $\displaystyle b_j$'s can produce the same coset. It remains to show that if $\displaystyle a b H = a' b'H$ then $\displaystyle a = a',b=b'$. Can you finish the last step? - Oct 10th 2008, 04:05 PMdori1123
So if $\displaystyle abH = a'b'H$, then $\displaystyle abh_1 = a'b'h_2$ for some $\displaystyle h_1, h_2 \in H$, and so $\displaystyle abh_1h_2^{-1}(b')^{-1}(a')^{-1} = 1 \iff h_1h_2^{-1} = 1, b(b')^{-1} = 1,$ and $\displaystyle a(a')^{-1} = 1$. So $\displaystyle a = a'$ and $\displaystyle b = b'$.

Is this correct? does it make sense? - Oct 11th 2008, 03:36 PMThePerfectHacker