Index satisfies intersection inequality

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Revision as of 13:07, 16 October 2008 by Vipul (talk | contribs) (New page: ==Statement== Suppose <math>G</math> is a group and <math>H, K</math> are subgroups of finite index in <math>G</math>. Then, we have: <math>[G:H \cap K] \le [G:H][G:K]</math>. (An analo...)
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Suppose G is a group and H, K are subgroups of finite index in G. Then, we have:

[G:H \cap K] \le [G:H][G:K].

(An analogous statement holds for subgroups of infinite index, provided we interpret the indices as infinite cardinals).

Related facts

Facts used

  1. Index satisfies transfer inequality: This states that if H, K \le G, then [K: H \cap K] \le [G:H].
  2. Index is multiplicative: This states that L \le K \le G, then [G:L] \le [G:K][K:L].


Given: A group G with subgroups H and K.

To prove: [G:H \cap K] \le [G:H][G:K].

Proof: By fact (1), we have:

[K:H \cap K] \le [G:H].

Setting L = H \cap K in fact (2) yields:

[G:H \cap K] = [G:K][K:H \cap K].

Combining these yields:

[G:H \cap K] \le [G:H][H:K]

as desired.