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- Thread starter murmillo
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Is there a question in there?

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Let G be a group of order 21. The Sylow 7-subgroup K must be normal, and the number of Sylow 3-subgroups is 1 or 7. Let x be a generator for K, and let y be a generator for one of the Sylow 3-subgrops H. Then x^7=1 and y^3=1, so H intersect K = {1} and therefore the product map HxK->G is injective. Since G has order 21, the product map is bijective.

OK, this following paragraph is where I'm stuck: Since K is normal, yxy^-1 is an element of K, a power of x, say x^i, where i is between 1 and 7. So the elements x and y satisfy the relations x^7=1, y^3=1, yx=(x^i)y. OK, this makes sense. But, says Artin, the relation y^3 restricts the possible exponents i:

x=(y^3)(x)(y^-3) = (y^2)(x^i)(y^-2) = y(x^(i^2))(y^-1)=x^(i^3)).

I don't understand the step after (y^2)(x^i)(y^-2). I understand that x is a generator of K, but is he saying that any generator g of K satisfies yxy^-1=g^i? Because we have

(y^2)(x^i)(y^-2) = y(y(x^i)y^-1)y^-1, and it seems that he's saying that y(x^i)y^-1 = x^(i^2)). So x goes to x^i under conjugation by y, but why does x^i go to x^(i^2)?

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VietDao29

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OK, this following paragraph is where I'm stuck: Since K is normal,yxy^-1 is an element of K, a power of x, say x^i, where i is between 1 and 7. So the elements x and y satisfy the relations x^7=1, y^3=1, yx=(x^i)y. OK, this makes sense. But, says Artin, the relation y^3 restricts the possible exponents i:

x=(y^3)(x)(y^-3) = (y^2)(x^i)(y^-2) = y(x^(i^2))(y^-1)=x^(i^3)).

Ok, so we know that:

[tex]yxy^{-1} = x ^ i \Rightarrow \prod_{k = 1} ^ i \left( yxy^{-1} \right) = \prod_{k = 1} ^ i (x ^ i)[/tex]

[tex]\Rightarrow y \left[ \prod_{k = 1} ^ {i - 1} \left( x y ^ {-1} y \right) \right] xy ^ {-1} = x ^ {i ^ 2}[/tex]

[tex]\Rightarrow y \left[ \prod_{k = 1} ^ {i - 1} x \right] xy ^ {-1} = x ^ {i ^ 2}[/tex]

[tex]\Rightarrow y x ^ i y ^ {-1} = x ^ {i ^ 2}[/tex]

You can do the same to obtain: [tex]y x ^ {i ^ 2} y ^ {-1} = x ^ {i ^ 3}[/tex]. :)

Hope you can get it. If you have any more questions, just don't hesitate to ask. :)

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