5
$\begingroup$

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

0

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

7

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
4

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?

  • 1
    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.2012-09-07
  • 0
    I take $\prod_{a\in G}a$ as $a_1.a_2...a_{2m}$. Is this remained ill-defined as Mark pinted?2012-09-07
  • 0
    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. :)2012-09-07
  • 1
    @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.2012-09-07
  • 0
    @MarcvanLeeuwen: So, should I edit my quoestion and add the group in non-abelian? I am asking just for sure dear sir.2012-09-07
  • 0
    I take $g$ as Hagen took so, $g^m\in H$ so, $g^{2k+1}H=H$ so, $(g^kH)^2.gH=H$ so, $gH=H$ which is a contradiction? Am I right?2012-09-07
  • 0
    @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
1

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.

  • 0
    Where did you use that $H$ is of odd order?2012-09-07
  • 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.2012-09-07
  • 0
    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.2012-09-07
  • 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.2012-09-07