Difference between revisions of "Index satisfies intersection inequality"

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Combining these yields:
Combining these yields:
<math>[G:H \cap K] \le [G:H][G:K]</math>
<math>[G:H \cap K] \le [G:K][G:H] = [G:H][G:K]</math>
as desired.
as desired.

Latest revision as of 16:57, 28 August 2017


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].


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:K][G:H] = [G:H][G:K]

as desired.