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Let $G$ be a finite group and $H\leq G$ be a subgroup of order odd such that $[G:H]=2$. Therefore the product of all elements in $G$ cannot belong to $H$.

I assume $|H|=m$ so $|G|=2m$. Since $[G:H]=2$ so $H\trianglelefteq G$ and that; half of the elements of the group are in $H$. Any Hints? Thanks.

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4 Answers 4

up vote 2 down vote accepted

For some fixed $g\in G-H$, we have $G = H \cup g H$ (disjoint). Then $(\prod_{a \in G} a)H = \prod_{a \in G} aH = \prod_{a \in G-H} aH=(gH)^m=gH$.


By the way, how do you define $\prod_{a \in G} a$ unambiguusly if $G$ is not necessarily abelian?

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If somewhat hard to read (you might have said the computation takes place in $G/H$), this shows the essence of the argument, +1. And although as you remark $\prod_{a \in G} a$ may be ill-defined, $(\prod_{a \in G} a)H$ is not. –  Marc van Leeuwen Sep 7 '12 at 9:40
    
I take $\prod_{a\in G}a$ as $a_1.a_2...a_{2m}$. Is this remained ill-defined as Mark pinted? –  Babak S. Sep 7 '12 at 10:00
    
It is not mentioned $G$ is abelian as in my old text, however, you are right. Indeed when $G$ is abelian we have nothing to be done. It would be meaningless. :) –  Babak S. Sep 7 '12 at 10:04
    
@BabakSorouh: To the contrary, if $G$ is non-Abelian the stronger statement is true that the product taken in any order will be outside $H$. The proof given here shows that. –  Marc van Leeuwen Sep 7 '12 at 10:06
    
@MarcvanLeeuwen: So, should I edit my quoestion and add the group in non-abelian? I am asking just for sure dear sir. –  Babak S. Sep 7 '12 at 10:15

HINTS:

  1. The product of elements of $H$ is in $H$.

  2. If $a,b\in G\setminus H$, $ab\in H$.

  3. $|G\setminus H|$ is odd.

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It's also useful to note that if $g\in G\setminus H$ and $h\in H$ then $gh = h^\prime g$ for some other element $h^\prime \in H$, so the order the product is taken doesn't matter. –  Sean Eberhard Sep 7 '12 at 11:27

Take the two different cosets of H in G as {H, gH}, g is not H.

Order of g is 2 in G/H. If gh1, gh2 are in gH; then their product is in H, since there is no element in common in H and gH; take h1 = 1 and; gh1*gh2 = h2 which is in H.

So the product of all elements of G is ghi*ghj..*hi*hj.. = k * ghi {for some k in H}

As H if odd order, we can rename product as gh0*(gh1*gh2*...gh2n) * k

As product of 2 elements of gH are in H, so the product (gh1*gh2*...gh2n) is in H.

So, the product is gh0*(gh1*gh2*...gh2n*k) is not in H.

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Where did you use that $H$ is of odd order? –  Marc van Leeuwen Sep 7 '12 at 9:33
2  
The question is about "the product of all elements of $G$"; while this forgets to specify an order, it does seem to mean that every element occurs only once, so I don't see what you mean by $g.g$. Also note that the product of all elements of the Klein $4$-group is the identity, which is in any index-$2$ subgroup $H$, so if you don't use that $|H|$ is odd, your proof cannot be correct. –  Marc van Leeuwen Sep 7 '12 at 10:04
    
Sorry Marc, I kept my brain some where, I understood the question as "show that there exists 2 elements in G whose product is not in H". I edited my answer accordingly. –  Ram Sep 7 '12 at 10:38
1  
"Order of g is 2" is false; the coset $gH$ has order 2 in $G/H$, which is a sensible statement only because $H$ is normal. Similarly, the computation "so the product (gh1*gh2*...gh2n) = l is in H" is incorrect, though the statement that product is in $H$ is correct. –  peoplepower Sep 7 '12 at 10:49

Consider the image of the product under the quotient map $G\to G/H\cong C_2$.

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