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April 8th, 2011, 12:56 PM   #1
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Question about Quotient Groups

Here is the question:
"Let H and K be normal subgroups of G such that [G : H] = 5 and [G : K] = 3.
Prove that for all g in G, (g^15) is in the intersection of H and K."
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I am told that I should start off this proof by showing (g^5) is in H by using the definition of quotient groups and Lagrange's Theorem.
Then, I assume I would use the same argument to show that (g^3) is in K?
Once I have done this, i want to combine both statements to show that (g^15) is in the intersection of H and K.
But I don't even know where to begin!

Thank you in advance for the assistance!
-J.R.B.
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April 8th, 2011, 02:23 PM   #2
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Re: Question about Quotient Groups

Start with the quotient group G/H. For any g in G, the coset (g^15)H = (gH)^15. But what is the order of the element gH in G/H? With that it mind, which coset must (gH)^15 equal?
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April 8th, 2011, 02:36 PM   #3
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Re: Question about Quotient Groups

From mtdim: "For any g in G, the coset (g^15)H = (gH)^15"

But the order of an element in G/H is 5, since [G:H] = |G/H| = 5.
(This must also imply that the order of an element in G/K is 3, since [G:K] = |G/K| = 3.)

Thus, since 5 divides 15, (gH)^15 = (g^15)H = ((g^5)^3)H = eH (e is the identity) so g^15 is in H.
Also, since 3 divides 15, (gK)^15 = (g^15)K = ((g^3)^5)K = eK (e is the identity) so g^15 is in K.

So if (g^15) is in H and K, (g^15) is in H intersect K? Is the proof done?
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April 8th, 2011, 03:00 PM   #4
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Re: Question about Quotient Groups

Not quite. You can't say ((g^5)^3)H = eH because you do not know the order of g in G. What you do know is that the order of any element of G/H must divide 5. So you can say: (g^15)H = (gH)^15 = eH because the order of gH must divide 15 (since it divides 5). Thus, g^15 must be in H. You can do a similar argument to show g^15 is in K.

The basic idea is that you don't really have any information about G, you only have information about G/H and G/K. You must look at this in terms of those quotient groups.
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