Isomorphism and group

[PotterAdmirer]

Hi.


I want to use the First Isomorphism Theorem to show that if H is a group and e €
H is identity element, then


H/{e}

≅​

H.


And this is my solution:


Let's look at function f : G --> H, which is given by f(x) = x^2. f is a homomorphism, because
f(xy) = (xy)^2 = f(x)f(y). f is surjective,
because if a €

H, then a = x^2 = a.​

Because e is identity element in H,
thus the kernel of f is all x €

G such that f(x) = x^2 = e.​

Elements that satisfy x^2 = e in fact are the elements with |x| ≤ 2.

ker f = e.


Thus, H/{e}

≅​

H.

I may have some errors in my solution ? But this is a best I can do.

 

[daon2]

Some problems I see: (xy)^2 and x^2*y^2 are not (in general) the same thing. Moreover, your map will often not be surjective. You put a 'G' into there, but you haven't quantified it. You also are using a norm on your group by stating |x|<=2., which doesn't make sense given the problem statement.

What tools do you have at your disposal? For example, this is trivial if one could use the First Isomorphism Theorem, with the map f being the identity map from H to H.

 

[PotterAdmirer]

Thank you for your response. This is really puzzling, but I will try again. Indeed, H is a group.

Let's look at function f : H --> H, which is given by f(x) = x^2. f is a homomorphism, because
f(xy) = (xy)^2. f is injective,
because if a €

H, then a = x^2 = a.​

Because e is identity element in H,
thus the kernel of f is all x €

H such that f(x) = x^2 = e.​

ker f = e.


Thus, H/{e}

≅​

H.


Well, I have for example http://en.wikipedia.org/wiki/Isomorphism_theorem

 

[daon2]

As I said f(x)=x^2 is often not a homomorphism. Why are you stuck on this one particular map? For example, let H be the dihedral group of order 8 (symmetries of a square). Let R be the element representing a rotation of pi/2 radians, and F a reflection over the y-axis. Assume we start with the following configuration of a square:

21
34

Then what does the element (RF)^2 do to the square? Well, (RF)^2=RFRF. This says flip, rotate, flip then rotate, you end up with:

21
34

which is what you started with! But (R^2)(F^2) simplifies to R^2 (since F^2=Identity transformation), which gives you

43
12

So f(RF) and f(R)f(F) are not the same things!

The reason this is NOT a homomorphism is because the group is not Abelian (commutative).

To show H and H/{e} are isomorphic, consider that have the same order. So to use the first isomorphism theorem, starting from H, you must produce an automorphism of H, i.e., you need a surjective map from H to H that has kernel {e}. Every group has at least two automorphisms (which someones coincide), they are the identity map f: x->x and the inverse-element map g: x -> x^{-1}.

 

[PotterAdmirer]

Thanks. I will try again.

Let's look at function f : H --> H


The map x -> x is an automorphism of H if and only if H is
abelian.


The map f: x->x and the inverse-element map g: x -> x^{-1}.


If H is an abelian group, then (xy)^-1
= y^-1 x^-1
= x^-1 y^-1


Now (xy)^-1 = x^-1 y^-1


= yx.
So, xy = yx


Hence,
H and H/{e} are isomorphic.

I'm afraid this is still incorrect.

 

[daon2]

That IS incorrect. You are not reading my post, you cannot assume H is abelian! I'm sorry, but I think you need direct help from your professor or face-to-face tutoring. This problem is supposed to be very straight-forward.