Answer to Question #137385 in Differential Equations for Tajesh

Question

Let G be a finite group and H be a subgroup of G with odd number of
elements such that (G:H) = 2. Show that the product of all elements of G (taken
in any order) does not belong to H.

Answer 1

[G:H]=2, so G is of even order =2n say and H is a normal subgroup of G.  Let $a\in G$ and o(a)=2.  Existence of a  is guaranteed by Cauchy’s Theorem. Then a doesn’t belong to H since the latter has odd order. Hence $G=H\cup aH.$  Now if we take product of all elements of G , then the product looks like product of elements of the form ah or h. Now

$ah=(aha^{-1})a$ and $ah_{i} ah_{j}=(ah_{i}a^{-1})( a^{2} h_{j}a^{-2} )a^2$ where each bracketed term belongs to H being normal. Hence the product can be written as $ha^{n}$. Hence the product is in H implies that $a^n \in H.$  But n is odd and order of a is 2. Hence $a^n =a.$ This implies $a\in H$ and hence we arrive at a contradiction.

Clarification through example:

$h_{1} h_2 (ah_2)$ $h_3 (ah_4)h_5 (ah_1)h_4 (ah_3 )(ah_5)=h_1h_2(ah_2 a^{-1})$ $(ah_3a^{-1}) (a^2h_4 a^{-2})(a^2h_5 a^{-2})(a^3h_1h_4 a^{-3})(a^4 ha^{-4 })(a^5h_5 a^{-5})a^5 \in H.$ We note each bracketed term is in H and hence $a\in H.$