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I have to show in a group $G$ of order $4k+2$ the product of the elements cannot be $1$.

What I got so far is that there exists a subgroup $H$ of index $2$ in $G$ (considering left regular representation on $G$).Then $G/H$ is abelian. Now consider the corresponding product of the cosets of $H$, i.e, $x_1x_2\dots x_nH$ where $n=4k+2$, we can arrange this product in such a way that this product becomes $x_1x_2\dots x_mH$ (after some renaming) where $m$ is also odd where each $x_i$ is of order $2$. I cannot proceed further.

Please help me, thank you.

egreg
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Via
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    "The product" is not well-defined unless you are talking abelian groups. – Thomas Andrews Sep 18 '14 at 21:45
  • It need not to be well defined bt not to be 1 – Via Sep 18 '14 at 21:47
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    It holds for any order of the product, though. Once you have a subgroup of index 2, $G/H$ has order 2, and the only group of order 2 is ${-1,1}$, up to isomorphism. The image in $G/H$ of a product of all elements will be $(-1)^{2k+1}1^{2+k1}=-1$ which can't be the image of $1$ under a homomorphism. – hmakholm left over Monica Sep 18 '14 at 21:48
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    Exactly U r right,my approach was wrong,i was taking it elementwise considring different order rather than cosets of H...thank U. – Via Sep 18 '14 at 21:53
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    My objection is to the usage of the term "the" in "the product." It implies there is a single value. I'd prefer a formulation like, "No product all the elements of $G$ is $1$." It's really a small nit, not a big deal. – Thomas Andrews Sep 18 '14 at 21:55
  • I understand it should no be 'the' before product,it depends on the order of the elements,U were right @ Thomas Andrews – Via Sep 18 '14 at 21:59
  • How did you prove the existence of $H$ ? – Belgi Sep 18 '14 at 22:03
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    Consider the multiplication map from G to G by a order 2 element in G,this will give u an on odd permutation – Via Sep 18 '14 at 22:05

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