# Math Help - Show that phi^(-1) is an isomorphism

1. ## Show that phi^(-1) is an isomorphism

Problem: Let $\phi:G \rightarrow G'$ be an isomorphism of groups. Show that $\phi^{-1}: G' \rightarrow G$ is an isomorphism.

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I said:

If $\phi:G \rightarrow G'$ is an isomorphism, $\phi$ is a bijection between G and G' that maps every x in G to an x' in G'. Since $\phi$ is a bijection, this map can be traced back from the x' in G' back to the x in G (that being the inverse of x'), using the inverse of the isomorphism function, $\phi^{-1}$.

How wrong am I?

2. ## Re: Show that phi^(-1) is an isomorphism

Originally Posted by tangibleLime
Problem: Let $\phi:G \rightarrow G'$ be an isomorphism of groups. Show that $\phi^{-1}: G' \rightarrow G$ is an isomorphism.

*********************

I said:

If $\phi:G \rightarrow G'$ is an isomorphism, $\phi$ is a bijection between G and G' that maps every x in G to an x' in G'. Since $\phi$ is a bijection, this map can be traced back from the x' in G' back to the x in G (that being the inverse of x'), using the inverse of the isomorphism function, $\phi^{-1}$.

How wrong am I?
You justified why [tex]\phi^{-1} exists, but why is it a homomorphism?

3. ## Re: Show that phi^(-1) is an isomorphism

hint: use the fact that φ is onto, and that φ^-1(φ(g)) = g.

4. ## Re: Show that phi^(-1) is an isomorphism

Hm... I was under the impression that an isomorphism is just a morphism that has an inverse. So a regular morphism (that's a thing, right?) can map something from X to Y but not necessarily from Y to X. An isomorphism, since it is a morphism with an inverse, can map from X to Y and then Y back to X. Is it enough to say that just by this definition of an isomorphism, if $\phi$ exists as the isomorphism, it's required inverse (required to actually be an isomorphism) is $\phi^{-1}$? So $\phi^{-1}(x) \in G' = x \in G$? Did I just inadvertently use the fact that $\phi$ is onto? Or do I need something more mathematically rigid, or am I just not going far enough?

5. ## Re: Show that phi^(-1) is an isomorphism

is is a fact that the inverse function of a bijective homomorphism is indeed a homomorphism itself, justifying the term isomorphism.

this fact, is what you are being asked to prove. that is if $\phi:G \to G'$ is a bijective homomorphism, you must show that the

function $\phi^{-1}:G' \to G$(which exists because φ is bijective) is, in fact, a homomorphism.

so you must show that for ANY $g',h' \in G', \phi^{-1}(g'h') = \phi^{-1}(g')\phi^{-1}(h')$

this is not immediately obvious from the fact that φ is a homomorphism.

6. ## Re: Show that phi^(-1) is an isomorphism

Hm, I think I understand what you are saying. So let me try again.

It is given that $\phi : G \rightarrow G'$ is an isomorphism. Need to prove that $\phi^{-1}: G' \rightarrow G$ is an isomorphism.

Since $\phi$ defines an isomorphism, it satisfies the homomorphism property:
$\phi(xy) = \phi(x)\phi(y) \hspace{4mm} \forall x,y \in G$

And $\phi$ maps values from G to G',
$\phi(x) = x'$, where $x \in G$ and $x' \in G'$.

Since $\phi$ is bijection, and therefore onto,
$\phi^{-1}(\phi(x)) = x$
$\Rightarrow \phi^{-1}(x') = x$

Using this fact with the homomorphism property,
$\phi^{-1}(\phi(xy)) = \phi^{-1}(\phi(x))\phi^{-1}(\phi(y))$
$\Rightarrow \phi^{-1}(x'y') = \phi^{-1}(x')\phi^{-1}(y')$
$\Rightarrow xy = xy$

And so $\phi^{-1}$ satisfies the homomorphism property, and is an isomorphism since it has an inverse $\phi$.

Am I close?

7. ## Re: Show that phi^(-1) is an isomorphism

Originally Posted by tangibleLime
Hm, I think I understand what you are saying. So let me try again.

It is given that $\phi : G \rightarrow G'$ is an isomorphism. Need to prove that $\phi^{-1}: G' \rightarrow G$ is an isomorphism.

Since $\phi$ defines an isomorphism, it satisfies the homomorphism property:
$\phi(xy) = \phi(x)\phi(y) \hspace{4mm} \forall x,y \in G$

And $\phi$ maps values from G to G',
$\phi(x) = x'$, where $x \in G$ and $x' \in G'$.

Since $\phi$ is bijection, and therefore onto,
$\phi^{-1}(\phi(x)) = x$
$\Rightarrow \phi^{-1}(x') = x$

Using this fact with the homomorphism property,
$\phi^{-1}(\phi(xy)) = \phi^{-1}(\phi(x))\phi^{-1}(\phi(y))$
$\Rightarrow \phi^{-1}(x'y') = \phi^{-1}(x')\phi^{-1}(y')$
$\Rightarrow xy = xy$

And so $\phi^{-1}$ satisfies the homomorphism property, and is an isomorphism since it has an inverse $\phi$.

Am I close?

you can't USE the homomorphism property of $\phi^{-1}$ until you prove it HAS it.

8. ## Re: Show that phi^(-1) is an isomorphism

To prove that $\phi^{-1}$ HAS the homomorphism property, is it enough to say...

$\phi(x)=x'$
And since $\phi$ is onto, $\phi^{-1}(x')=x$.

Therefore, there exist one-to-one correspondences between elements of G and G',
$x \leftrightarrow x'$
$y \leftrightarrow y'$

Therefore, G and G' have the same cardinality and the following one-to-one correspondence exists:
$xy \leftrightarrow x'y'$

Which can be expressed as the homomorphism property $\phi^{-1}(x'y')=\phi^{-1}(x')\phi^{-1}(y')$.

I feel like I'm straying off in the wrong direction

9. ## Re: Show that phi^(-1) is an isomorphism

what we would LIKE to prove is that:

$\phi^{-1}(x'y') = \phi^{-1}(x')\phi^{-1}(y')\ \forall x',y' \in G'$

but we don't know that this is true. in fact, we know very little about the behavior of $\phi^{-1}$.

what we DO know, is about the behavior of $\phi$.

so what we need to do is somehow get to where we're working with $\phi$, so we can use IT's homomorphism property.

φ is onto, so if we have x' in G', we know that x' = φ(x) for some x in G. similarly y' = φ(y), for some y in G.

so:

$\phi^{-1}(x'y') = \phi^{-1}(\phi(x)\phi(y))$

now, because φ is a homomorphism, we know that φ(x)φ(y) = φ(xy). thus:

$\phi^{-1}(\phi(x)\phi(y)) = \phi^{-1}(\phi(xy))$

note that we haven't done ANYTHING with $\phi^{-1}$ yet.

now, because φ is a bijection, and φ^-1 is its inverse function (not homomorphism, because we don't know that yet):

$\phi^{-1}(\phi(xy)) = xy$

but what are x and y? they are the elements φ maps to x',y':

φ(x) = x'
φ(y) = y', so

$x = \phi^{-1}(\phi(x)) = \phi^{-1}(x')$ and similarly $y = \phi^{-1}(\phi(y)) = \phi^{-1}(y')$, so that:

$xy = \phi^{-1}(x')\phi^{-1}(y')$

and THAT shows $\phi^{-1}$ is a homomorphism, by transitivity of equality.

10. ## Re: Show that phi^(-1) is an isomorphism

Thanks, I don't think I ever would have gotten that. Though I understood everything you did, and why you did it. Hopefully come exam time, I'll be able to think creatively enough to pull something like this out.

11. ## Re: Show that phi^(-1) is an isomorphism

a general problem solving technique is this:

you want to get from point A to point B. you don't know how, the theorems you have don't quite fit, there's some "missing piece".

you do however, have tools that tell you how to get from point C to point D, they're the things that you use over and over again.

so: look for a path between the context A and B are in, and the context C and D are in (this process is called "translating a problem").

then use your path from C to D, and "translate back":

A-- xxxx -->B (broken path)

A--->C--->D--->B (problem solved!)

here the "broken path" was: homomorphic property of φ^-1. we didn't even know how to try to compute stuff in G'.

but since we had a bijection with φ and φ^-1, we first use the bijection (the function part, which we knew) to go "back to G".

now in G, we had the desired homomorphism property, so no problem there. so we get a nice equation we can use.

then we "translate" back to G', using our bijection property.

so: when you see the words "iso-something" or "equivalence" or "1-1 correspondence" or "bijection" or anything with double arrowheads: <---->

you should think in your mind: translation dictionary, can be used to put problems in different context.

one of the reasons cyclic groups are so important for group theory is this:

if we can somehow put a group theory problem in terms of cyclic somethings, we can turn the cyclic group information into

information about integers. well, integers are easy, it's just basic (sometimes modular) arithmetic.

translate the integer information back into cyclic group information, and then use it for a general group.

if you look closely enough, this is exactly what is done in every proof of lagrange's theoem.