# Isomorphism (Fraleigh 7th: Section 5)

1. Aug 20, 2009

### jeff1evesque

Directions:
Let $$\phi: G \rightarrow G'$$ be an isomorphism of a group $$<G, *>$$ with a group $$<G', *'>$$. Write out a proof to convince a skeptic of the intuitive clear statement.

Problem:
41.) If H is a subgroup of G, then $$\phi[H] = {\phi(h)| h \in H}$$ is a subgroup of $$G'$$. That is, an isomorphism carries subgroups into subgroups.

Thoughts on the problem:
I am not sure what is mean by "an isomorphism carries subgroups into subgroups." Nevertheless, I went to get help to draw up a proof for this problem. After coming up with a proof, I reviewed it a little later, and was a little skeptical about the homomorphism aspect (shown below).

Proof:
Let $$x, y \in \phi(H).$$
Now, Suppose $$\exists a, b \in H: x = \phi(a), y = \phi(b).$$
Therefore,
$$xy = \phi(a) \phi(b) = \phi(ab) \in \phi(H).$$ (#1)
Also, $$e = \phi(e) \in \phi(H),$$ (since the identity is in G, the subgroup H must have the same identity). (#2)
Finally, $$x^{-1} = [\phi(a)]^{-1} = \phi(a^{-1})$$ shows there is an inverse element in the subgroup H. (#3)

Questions:
In equation (#1), can we conclude that $$\phi(a) \phi(b) = \phi(ab)?$$ It seems kind of vague to me but all I've gathered from the problem is that our function $$\phi$$ takes the group G with its binary operator * to another group G' with an operator *'. Neither operators (*, *') are defined, so how can we conclude $$\phi(a) \phi(b) = \phi(ab)?$$

In equation (#2), and equation (#3) the reasoning seems fairly straightfoward. In these two equations we've shown that the identity $$e \in H$$ is also in the group G'. Similarly we've shown that the same inverse in H (which is a subgroup of G), is also found in the group G'.

So what I'm really puzzled about is how the first equation (equation (#1)), contains/proves the homomorphism property (to obtain the closure condition). We've shown that $$xy = \phi(ab) \in \phi(H)$$ fulfills closure, this is useful since we were trying to prove the subgroup condition. However, in doing so we had to show the homomorphism property to get there. To me the whole proof is pretty good except I just can accept that $$\phi(a) \phi(b) = \phi(ab)$$.

Thanks,

JL

Last edited: Aug 20, 2009
2. Aug 20, 2009

### HallsofIvy

You understand, don't you, that the words "that is" means that they are just restating what was said before: "if H is a subgroup of G then $\phi(H)$ is a subgroup of G'".

Yes, we can. That is part of the definition of "homomophism" and an "isomorphism" is a special kid of homomorphism.

3. Aug 20, 2009

### jeff1evesque

In my problem I had,
$$\phi(a) \phi(b) = \phi(ab)$$

$$\phi(a) @ \phi(b)$$

Would it be rewritten by the following,

$$\phi(a) @ \phi(b) = \phi(a @ b)?$$ Would this too would fulfill the homomorphism property (for whatever kind of problem it may be)?

I guess I just wanted to know if we can move a, b (from separate paranthesis) into the common paranthesis for all cases- that is for all operators (or all functions $$\phi$$)?
$$\phi(a) \phi(b) = \phi(ab)?$$
$$\phi(a) @ \phi(b) = \phi(a @ b)?$$
$$\phi(a) ! \phi(b) = \phi(a ! b)?$$
and so on...

Thanks.

Last edited: Aug 20, 2009
4. Aug 21, 2009

### jeff1evesque

Given our problem,
$$\phi : G \rightarrow G'$$ is an isomorphism of the group <G, *> with a group <G', *'>

So $$\phi(x*y) = \phi(x) *' \phi(y)$$ for all $$x, y \in S$$ (homomorphism property is included since we're told it is isomorphic)

But how can $$\phi(xy) = \phi(x) \phi(y)$$ since each binary structure (<G, *>, <G', *'>) has binary operator *, and *' respectively. $$\phi(xy)$$ has no binary operator, and thus is not part of the binary structure in the domain and is not part of the binary structure of the codomain. If $$\phi(xy)$$ has no binary operator, then can we assign a value to it- in particular $$\phi(x) \phi(y)$$?

5. Aug 21, 2009

### Hurkyl

Staff Emeritus
if x,y are in the group (G,*), then xy is shorthand for x*y

6. Aug 21, 2009

### jeff1evesque

Sorry for silly questions, but I think I get it (notations). If x,y are in the group (G,*), then if we apply the function $$\phi$$ we get $$\phi(xy) = \phi(x*y) = \phi(x) * \phi(y)$$

So $$\phi(xy)$$ is really shorthand for $$\phi(x*y)$$, and $$\phi(x)\phi(y)$$ is really short for $$\phi(x)* \phi(y)$$.

Last edited: Aug 22, 2009
7. Aug 22, 2009

### jeff1evesque

Although in this proof we showed closure, the identity, and the inverse exists, it seems to me all that we did was prove that H is a subgroup of G. How do I conclude $$\phi[H] = {\phi(h)| h \in H}$$ is a subgroup of $$G'$$? Is this already implied in the proof?

8. Aug 22, 2009

### VeeEight

It was a given assumption that H is a subgroup of G. You checked that phi(H) is a subgroup of G' in your first post under 'proof' (although you should clean it up a bit)

9. Aug 23, 2009

### Tobias Funke

Just a tiny correction for what may have just been a typo, but this should read:

So $$\phi(xy)$$ is really shorthand for $$\phi(x*y)$$, and $$\phi(x)\phi(y)$$ is really short for $$\phi(x)*' \phi(y)$$

Also, I learned from this book so I know that he doesn't emphasize the shorter test to determine whether a subset is a subgroup. He does mention it and then asks you to prove it (problem 45 on that same page), so if you're still uncomfortable with your proof you may want to try it that way.