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 Post subject: Does every element have a square root problemPosted: Sun, 29 Jan 2012 19:15:23 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
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Prove it is true for every group in G or disprove by counterexample

For every x in G there is some y in G such that x=y^2 (This is the same as saying every element in G has a square root)

Now since x=y^2, am i trying to prove that also y=x^2?
If this is the case i am having lots of trouble

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 Post subject: Re: Does every element have a square root problemPosted: Sun, 29 Jan 2012 20:06:58 UTC
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Joined: Wed, 21 May 2003 04:27:18 UTC
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DgrayMan wrote:
Prove it is true for every group in G or disprove by counterexample

For every x in G there is some y in G such that x=y^2 (This is the same as saying every element in G has a square root)

Now since x=y^2, am i trying to prove that also y=x^2?
If this is the case i am having lots of trouble

I'm thinking the answer is negative. Couldn't you use the set of 2x2 (invertible) matrices under multiplication, for example?
If you use as your field of scalars, I believe (although I haven't worked this out) that you can prove that the diagonal matrix with -1 for each diagonal entry would not have a square root (sorry, I'm rusty on my LaTeX).

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"Mathematicians are like lovers. Grant a mathematician the least principle, and he will draw from it a consequence which you must also grant him, and from this consequence another." Bernard Le Bovier Fontenelle (1657-1757)

"In great mathematics there is a very high degree of unexpectedness, combined with inevitability and economy."
G.H. Hardy (1877-1947)

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 Post subject: Re: Does every element have a square root problemPosted: Sun, 29 Jan 2012 22:12:18 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12071
Location: Austin, TX
Justin wrote:
DgrayMan wrote:
Prove it is true for every group in G or disprove by counterexample

For every x in G there is some y in G such that x=y^2 (This is the same as saying every element in G has a square root)

Now since x=y^2, am i trying to prove that also y=x^2?
If this is the case i am having lots of trouble

I'm thinking the answer is negative. Couldn't you use the set of 2x2 (invertible) matrices under multiplication, for example?
If you use as your field of scalars, I believe (although I haven't worked this out) that you can prove that the diagonal matrix with -1 for each diagonal entry would not have a square root (sorry, I'm rusty on my LaTeX).

Or just take ....

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 Post subject: Re: Does every element have a square root problemPosted: Mon, 30 Jan 2012 00:37:22 UTC
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Justin wrote:
DgrayMan wrote:
Prove it is true for every group in G or disprove by counterexample

For every x in G there is some y in G such that x=y^2 (This is the same as saying every element in G has a square root)

Now since x=y^2, am i trying to prove that also y=x^2?
If this is the case i am having lots of trouble

I'm thinking the answer is negative. Couldn't you use the set of 2x2 (invertible) matrices under multiplication, for example?
If you use as your field of scalars, I believe (although I haven't worked this out) that you can prove that the diagonal matrix with -1 for each diagonal entry would not have a square root (sorry, I'm rusty on my LaTeX).

No, that would not work:.

Instead, consider the multiplicative group , or any finite group of even order.

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 Post subject: Re: Does every element have a square root problemPosted: Mon, 30 Jan 2012 00:43:41 UTC
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outermeasure wrote:
Justin wrote:
DgrayMan wrote:
Prove it is true for every group in G or disprove by counterexample

For every x in G there is some y in G such that x=y^2 (This is the same as saying every element in G has a square root)

Now since x=y^2, am i trying to prove that also y=x^2?
If this is the case i am having lots of trouble

I'm thinking the answer is negative. Couldn't you use the set of 2x2 (invertible) matrices under multiplication, for example?
If you use as your field of scalars, I believe (although I haven't worked this out) that you can prove that the diagonal matrix with -1 for each diagonal entry would not have a square root (sorry, I'm rusty on my LaTeX).

No, that would not work:.

Instead, consider the multiplicative group , or any finite group of even order.

Oh right, my example was a bit off, I mean , sorry. Brain has been on fields and reciprocity laws, but I forget to omit the 0, and the additive group structure is definitely not a problem.

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 Post subject: Re: Does every element have a square root problemPosted: Mon, 30 Jan 2012 00:54:32 UTC
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Maybe I just don't understand the question. Consider the group of all non-zero complex numbers under multiplication. Every number has a square root, but the square root of the square root is not the original number.

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 Post subject: Re: Does every element have a square root problemPosted: Mon, 30 Jan 2012 00:55:42 UTC
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mathematic wrote:
Maybe I just don't understand the question. Consider the group of all non-zero complex numbers under multiplication. Every number has a square root, but the square root of the square root is not the original number.

Eh? He wants to know if every group has the property that there is a square root relative to the group operation, and that's clearly false.

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 Post subject: Re: Does every element have a square root problemPosted: Mon, 30 Jan 2012 04:53:51 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
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so with my original assumption. if x=y^2, there isn't always a x^2=y?
So can i find a y which y^2=x but x^2 does not equal y, to show this doesn't work?
Or am i misinterpreting it completely?

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 Post subject: Re: Does every element have a square root problemPosted: Mon, 30 Jan 2012 04:55:13 UTC
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DgrayMan wrote:
so with my original assumption. if x=y^2, there isn't always a x^2=y?
So can i find a y which y^2=x but x^2 does not equal y, to show this doesn't work?
Or am i misinterpreting it completely?

Again, NO. Check out the examples we've given you so far. Any group of even order should immediately convince you that this is a ridiculous notion.

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 Post subject: Re: Does every element have a square root problemPosted: Mon, 30 Jan 2012 07:16:58 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
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i guess i have o look up matrix rules again im so confused now.

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 Post subject: Re: Does every element have a square root problemPosted: Mon, 30 Jan 2012 14:35:25 UTC
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DgrayMan wrote:
i guess i have o look up matrix rules again im so confused now.

Why do you need to check on matrix rules for this question? Are you familiar at all with the theory of finite groups?

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 Post subject: Re: Does every element have a square root problemPosted: Mon, 30 Jan 2012 15:01:46 UTC
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outermeasure wrote:
Justin wrote:
DgrayMan wrote:
Prove it is true for every group in G or disprove by counterexample

For every x in G there is some y in G such that x=y^2 (This is the same as saying every element in G has a square root)

Now since x=y^2, am i trying to prove that also y=x^2?
If this is the case i am having lots of trouble

I'm thinking the answer is negative. Couldn't you use the set of 2x2 (invertible) matrices under multiplication, for example?
If you use as your field of scalars, I believe (although I haven't worked this out) that you can prove that the diagonal matrix with -1 for each diagonal entry would not have a square root (sorry, I'm rusty on my LaTeX).

No, that would not work:.

Instead, consider the multiplicative group , or any finite group of even order.

Oops, guess I wasn't as thorough as I thought...

So we could even use something as basic as eh? I should have thought of that.

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"Mathematicians are like lovers. Grant a mathematician the least principle, and he will draw from it a consequence which you must also grant him, and from this consequence another." Bernard Le Bovier Fontenelle (1657-1757)

"In great mathematics there is a very high degree of unexpectedness, combined with inevitability and economy."
G.H. Hardy (1877-1947)

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 Post subject: Re: Does every element have a square root problemPosted: Mon, 30 Jan 2012 15:03:15 UTC
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DgrayMan wrote:
i guess i have o look up matrix rules again im so confused now.

To original poster: Ignore my example, it doesn't work...although I do remember something from linear algebra about not every square matrix having a square root. I think that result is much harder than the problem intended.

Here is how complicated it can get with matrices: http://en.wikipedia.org/wiki/Square_root_of_a_matrix

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"Mathematicians are like lovers. Grant a mathematician the least principle, and he will draw from it a consequence which you must also grant him, and from this consequence another." Bernard Le Bovier Fontenelle (1657-1757)

"In great mathematics there is a very high degree of unexpectedness, combined with inevitability and economy."
G.H. Hardy (1877-1947)

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 Post subject: Re: Does every element have a square root problemPosted: Mon, 30 Jan 2012 15:11:09 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
whats an example of a 2x2 matrix that does not have a square root?

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 Post subject: Re: Does every element have a square root problemPosted: Mon, 30 Jan 2012 15:35:04 UTC
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Justin wrote:
DgrayMan wrote:
i guess i have o look up matrix rules again im so confused now.

To original poster: Ignore my example, it doesn't work...although I do remember something from linear algebra about not every square matrix having a square root. I think that result is much harder than the problem intended.

Here is how complicated it can get with matrices: http://en.wikipedia.org/wiki/Square_root_of_a_matrix

Justin: if you want a matrix (over say) without a square root, just get one with a negative determinant like .

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