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# Answer to Question #17574 in Abstract Algebra for steven peters

Question #17574
Let K<=H<=G (subgroups). Prove that if index (G:K) is finite, then indices (H:K) and (G:H) are also finite.
Expert's answer
Let B be a group and A<B be a subgroup.
Say that two elements x,y from B are equivalent x ~ yif x = ay for some a from A.
Each equivalency class is called adjacent class, and theset of all such classes (i.e. factor space) is called the set of adjacency
classes and denoted by B/A.
Then the index (B:A) is the number of elements in B/A.

Thus each adjacency class of B/A has the form A*y for some y from B.

Suppose now that K < H < G.

1) Let K*x, (xfrom H), be any adjacency class of H/K.
Since x belogns to G as well, we see that K8x is also anadjacency class of G/K, whence number (H:K) of classes in H/K is not greater than the number (G:K) of classes in G/K, and so
(H:K) <= (G:K) < infinity, so (H:K) is finite.

2) Let H*y, (y from G), be any adjacency class of G/H.
Since H is a dijoint union of (H:K) adjacency classes
H*x_1, H*x_2, ...
we see that Hy is also a disjoint union of (H:K) classes
H*x_1*y, H*x_2*y, ...

As (G:K), and (H:K) are finite we obtain that
(G:H) = (G:K) / (H:K)
and this number is also finite.

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