Difference between revisions of "Index satisfies intersection inequality"

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(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...)
 
(Facts used)
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# [[uses::Index satisfies transfer inequality]]: This states that if <math>H, K \le G</math>, then <math>[K: H \cap K] \le [G:H]</math>.
 
# [[uses::Index satisfies transfer inequality]]: This states that if <math>H, K \le G</math>, then <math>[K: H \cap K] \le [G:H]</math>.
# [[uses::Index is multiplicative]]: This states that <math>L \le K \le G</math>, then <math>[G:L] \le [G:K][K:L]</math>.
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# [[uses::Index is multiplicative]]: This states that <math>L \le K \le G</math>, then <math>[G:L] = [G:K][K:L]</math>.
  
 
==Proof==
 
==Proof==

Revision as of 23:40, 12 May 2010

Statement

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] = [G:K][K:L].

Proof

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.