if (x^3)ax=a' then a has a cube root.

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

Let set G be integers with x and a.
if (x^3)ax=a' then a has a cube root.
(Hint: show that xax is a cube root of a')

Ok, since the set is integers that means that (x^3)ax can be in any order and still equal a', correct??

Now then, what exactly am i trying to find? is it that a=( ? )^3 ??

Then is the hint meaning that i should find a'=(xax)^3 ?? I have other problems of this nature, does anyone have a simple example to help me understand this?

outermeasure · **Posted:** Mon, 30 Jan 2012 16:34:09 UTC

DgrayMan wrote:

Let set G be integers with x and a.
if (x^3)ax=a' then a has a cube root.
(Hint: show that xax is a cube root of a')

Ok, since the set is integers that means that (x^3)ax can be in any order and still equal a', correct??

Now then, what exactly am i trying to find? is it that a=( ? )^3 ??

Then is the hint meaning that i should find a'=(xax)^3 ?? I have other problems of this nature, does anyone have a simple example to help me understand this?

What is a'? What is "integers with x and a"?

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

a' means a inverse. and x and a are in G, which is a set of integers

Shadow · **Posted:** Mon, 30 Jan 2012 18:07:32 UTC

DgrayMan wrote:

Let set G be integers with x and a.
if (x^3)ax=a' then a has a cube root.
(Hint: show that xax is a cube root of a')

Ok, since the set is integers that means that (x^3)ax can be in any order and still equal a', correct??

Now then, what exactly am i trying to find? is it that a=( ? )^3 ??

Then is the hint meaning that i should find a'=(xax)^3 ?? I have other problems of this nature, does anyone have a simple example to help me understand this?

I think you'd best get your notation fixed, as it stands you're asking to show that an integer,

has a cube root if there is another integer

such that $ax^4={1\over a}$ , which I guess is true, but only because these conditions imply that $a=\pm 1$ and $x=\pm 1$ .

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

I wrote the problem wrong sorry its (x^2)ac=a' . its hard to write it here on the forum so that's the best i can do. the set is {G,*} the problem is x squared ac = a'

Shadow · **Posted:** Mon, 30 Jan 2012 18:14:40 UTC

DgrayMan wrote:

I wrote the problem wrong sorry its (x^2)ac=a' . its hard to write it here on the forum so that's the best i can do. the set is {G,*} the problem is x squared ac = a'

If you're having problems typesetting equations, check out our $\LaTeX$ help board at this link.

There's still the problem of what is c in this expression now? Can you pick it? Is it fixed? Is it supposed to be an integer?

Shadow · **Posted:** Mon, 30 Jan 2012 18:26:19 UTC

Wait, is this supposed to be a group theory question? If so you need to specify the group operation, since you appear to be writing it multiplicatively. Since your last post was similar and on group theory I'm assuming this is the case and moving this to Algebra and Number Theory, if this really is just a basic algebra question, you should tell me to move it back.

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

ugh. i mean (x^2)ax=a'. no c in there.

Shadow · **Posted:** Mon, 30 Jan 2012 18:36:02 UTC

DgrayMan wrote:

ugh. i mean (x^2)ax=a'. no c in there.

Again, my question is as to whether or not this is a group theory question, and what the group operation is. Integers under addition doesn't make sense, because then it's false just by other facts about integers.

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

yes its group theory, the operation is just any operation denoted by *.

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

im gonna scan the page this problem is on and post it so this confusion will end.

Shadow · **Posted:** Mon, 30 Jan 2012 18:54:14 UTC

DgrayMan wrote:

im gonna scan the page this problem is on and post it so this confusion will end.

That's probably best. It's common to write

for a general group operation, but you specified that the set was the integers, so generally this means you also want addition to be your group operation.

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

Question #7 is the one I'm working on

Shadow · **Posted:** Mon, 30 Jan 2012 22:13:54 UTC

DgrayMan wrote:

Question #7 is the one I'm working on

Oh, then this is easy.

(xax)^3=xax^2ax^2ax

, now using the fact that the group law is associative, and that $x^2ax=a^{-1}$ we get:

$xax^2ax^2ax=(x^{-1})(x^2axa)(a^{-1}xa)(x^2axa)$ $(a^{-1})=x^{-1}ea^{-1}xaea^{-1}=x^{-1}a^{-1}x$

On the other hand, we know:

$(xax)^3=xax^2ax^2ax=x(ax^2ax)x(ax^2ax)=xexe=x^2\quad (*)$

So we get:

$x^2=x^{-1}a^{-1}x\iff x= (ax)^{-1}\iff x^{-1}=ax$ , but then $a=x^{-2}$ so $a^{-1}=x^2=(*)$ , hence xax

is a cube root for $a^{-1}$ .

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

you gotta be kidding me?!? :shock:

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if (x^3)ax=a' then a has a cube root.

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