Does every element have a square root problem

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

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

Justin · **Joined:** Wed, 21 May 2003 04:27:18 UTC **Posts:** 988

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 $\mathbb{R}$ 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).

Shadow · **Posted:** Sun, 29 Jan 2012 22:12:18 UTC

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 $\mathbb{R}$ 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 $G=\mathbb{Z}/3\mathbb{Z}$ ....

outermeasure · **Posted:** Mon, 30 Jan 2012 00:37:22 UTC

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 $\mathbb{R}$ 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: $\begin{pmatrix}0&-1\\1&0\end{pmatrix}^2=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ .

Instead, consider the multiplicative group $\mathbb{R}-\{0\}$ , or any finite group of even order.

Shadow · **Posted:** Mon, 30 Jan 2012 00:43:41 UTC

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 $\mathbb{R}$ 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: $\begin{pmatrix}0&-1\\1&0\end{pmatrix}^2=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ .

Instead, consider the multiplicative group $\mathbb{R}-\{0\}$ , or any finite group of even order.

Oh right, my example was a bit off, I mean $(\mathbb{Z}/3\mathbb{Z})^\times$ , 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.

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 321

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.

Shadow · **Posted:** Mon, 30 Jan 2012 00:55:42 UTC

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.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

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?

Shadow · **Posted:** Mon, 30 Jan 2012 04:55:13 UTC

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.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

i guess i have o look up matrix rules again

im so confused now.

Shadow · **Posted:** Mon, 30 Jan 2012 14:35:25 UTC

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?

Justin · **Joined:** Wed, 21 May 2003 04:27:18 UTC **Posts:** 988

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 $\mathbb{R}$ 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: $\begin{pmatrix}0&-1\\1&0\end{pmatrix}^2=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ .

Instead, consider the multiplicative group $\mathbb{R}-\{0\}$ , or any finite group of even order.

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

So we could even use something as basic as $\mathbb{Z}_2$ eh? I should have thought of that.

Justin · **Joined:** Wed, 21 May 2003 04:27:18 UTC **Posts:** 988

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

DgrayMan · **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?

Shadow · **Posted:** Mon, 30 Jan 2012 15:35:04 UTC

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 $\mathbb{R}$ say) without a square root, just get one with a negative determinant like $\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$ .

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Does every element have a square root problem

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