Proof of uniqueness of square root

[SticksandStones]

Homework Statement

Let G be a finite group in which every element has a square root. That is, for each x$$\epsilon G$$, there exists $$y \epsilon G$$ such that $$\(y^2=x.\)$$Prove that every element in G has a unique square root.

The Attempt at a Solution

Proof: Assume not. Let k be the order of G. Let $$\(y_0, y_1, y_2,\text{...},\)\(y_k\)$$be all the elements of G. For $$\(y_i{}^2\)= \(x_i\)$$it can be said
that $$\(y_i\)*\(y_i\)=\(x_i\)$$ and that $$\(y_i\)$$ is the square root of $$\(x_i\)$$. If $$\(x_i\)$$ has more than one square root, then there would be two unique
elements $$\(y_i\)$$ and $$\(y_j\)$$ with $$i \neq j$$ such that $$\(y_i\)*\(y_i\) = \(x_i\) =\(y_j\)*\(y_j\) = \(x_j\)$$. However, if $$\(x_i\)$$ is the identity
element, then $$\(y_i\)$$ must also be the identity element. However, $$\(x_i\) = \(y_j\)*\(y_j\)$$ and thus $$\(y_j\)$$ must also be the identity element. But
$$\(y_i\)$$ and $$\(y_j\)$$ are distinct elements of G, and thus a contradiction is reached.



I feel like I'm only proving that the identity element can not have two distinct square roots. Am I on the right path?

 

[SticksandStones]

Maybe this might be better: let y_i and y_j must have unique inverses. Let these be z_i and z_j respectively.
y_i*y_i = x_i
y_i*y_i = y_j * y_j
y_i*y_i*z_i = y_j*y_j*z_i
y_i = y_j*y_j*z_i
z_i*y_i = z_i*y_j*y_j*z_i
1 = z_i*y_j*y_j*z_i

But this can only be true if z_i*y_j = 1 and y_j*z_i = 1, which can only occur if z_i and y_j are inverses. However, inverses are unique, and if z_i and y_i are inverses and z_i and y_j are inverses, then y_i must be y_j. However, this contradicts the original statement of y_i and y_j being distinct elements of G.

 

[Dick]

Think of it this way. Let f(x)=x^2. f maps G to G. Since every element has a square root, f is onto. G is FINITE (very important). Can a map from a finite set G->G be onto if it's not one-to-one?

 

[SticksandStones]

Yeah, but I believe I'm supposed to be working only with the axioms of groups (associativity, inverse, identity element, law of exponents and the such).

Edit: It's an Intro Abstract Algebra course.

 

[Dick]

Yeah, but I believe I'm supposed to be working only with the axioms of groups (associativity, inverse, identity element, law of exponents and the such).

You also have to use the properties of a set being finite. Or it's not true. Take the group of complex numbers of magnitude 1 under multiplication. Every element has a square root. But the square root is not unique, every number has two.

 

[SticksandStones]

You also have to use the properties of a set being finite. Or it's not true. Take the group of complex numbers of magnitude 1 under multiplication. Every element has a square root. But the square root is not unique, every number has two.

Ok, so if G has order k then...

Let I be the set of all y_i 0=<i=<k
and let J be the set of all x_i 0<i<k

f(x) = x^2 maps I->J so that f(y_i) = x_i

If some x_i has more than one square root, such as y_a and y_b, then there are only k-2 remaining elements in I to map to k-1 elements in J. So some element x_i could not have a square root unless y_c*y_c = x_c and x_d.

Is this correct?

 

[Dick]

Right. If one element has two square roots, then there are only k-2 candidates for the other k-1 elements that need square roots. There aren't enough to go around.
I would stop talking after "So some element x_i could not have a square root". I'm not sure what the "unless" is about.

 

[SticksandStones]

Right. If one element has two square roots, then there are only k-2 candidates for the other k-1 elements that need square roots. There aren't enough to go around.
I would stop talking after "So some element x_i could not have a square root". I'm not sure what the "unless" is about.

Sorry, it's late. I was trying to say that unless some number squared could somehow have two values at the same time that it would be impossible for every element to have a square root. I'll go with what you said and just stop talking there.


And now I feel stupid for not recognizing this earlier. Thanks for the help man!