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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.

4 Answers 4

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

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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.

  • 0
    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.2012-09-07
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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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    @MarcvanLeeuwen The question has been copied incorrectly. This is exercise 72 in Gallian's Contemporary Abstract Algebra in Chapter 9 where it is stated as"...(taken in any order)..."2016-03-23
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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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    "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.2012-09-07