# Two Group homomorphism proofs

1. Feb 5, 2013

### CAF123

1. The problem statement, all variables and given/known data
Let $\theta : G \mapsto H$ be a group homomorphism.
A) Show that $\theta$ is injective $\iff$ $\text{Ker}\theta = \left\{e\right\}$

B) If $\theta$ is injective, show that $G \cong Im \theta ≤ H$.

3. The attempt at a solution

A)The right implication is fine. Briefly, if $\theta$ is a homomorphism, then the identity in G is always sent to the identity in H. Since $\theta$ is injective, no other element maps there. I am not so sure about the left implication, but what I tried was:
Consider $\theta (g_1) = h_1$ and $\theta (g_2) = h_1.$ Want to show that $g_1 = g_2$ using the fact that $\text{Ker}\theta =$identity.
Then $\theta (g_1 e_G) = \theta (g_1) e_H = h_1 = \theta(g_2)e_H$ so $\theta(g_1) = \theta(g_2) (1)$. But I realise this is a trivial result. From here I want to show that $g_1 = g_2$. We were given a hint to use inverses more explicitly, but I can't see it right now.

B)I think I have this one, I just would like someone to check over it to make sure I haven't overlooked anything. So prove that $G \cong Im \theta$

This means $G$ is isomorphic to $Im \theta$ and so we have a bijective homomorphism. It is given that $\theta$ is injective, so we need only show surjectivity.
But simply, by defintion of $Im \theta$, it is surjective.

Now prove $Im \theta \leq H$, so use test for subgroup.
i)Non empty: we are dealing with a homomorphism, so the identity in g is always mapped to identity in H, so the image is never nonempty.

ii)Closure: Let $h_1, h_2 \in Im \theta$. For some $g_1, g_2 \in G, g_1 \neq g_2$, we have $\theta(g_1) = h_1$ and $\theta(g_2) = h_2$. Then $\theta(g_1)\theta(g_2) = h_1 h_2 = \theta(g_1 g_2).$ We found such an element that maps to $h_1 h_2$ so closure is satisfied.

iii) Let $\ell = h_1 h_2 \Rightarrow \ell h_2^{-1} = h_1 \Rightarrow \ell h_2^{-1} h_1^{-1} = e \Rightarrow \ell (h_1 h_2)^{-1} = e.$ So we found an element such that when it is smashed with another element we get $e \in Im \theta$, so closed under inverse.

Is this okay? If so, it is really just the left implication of A) that I need a hint for.

Many thanks

2. Feb 5, 2013

### Kreizhn

For 1, what happens to the element $g_1 g_2^{-1}$ under the homomorphism?

For 2, your proof is correct though you already know e is in the image as identity is preserved under the morphism. This is also a special case of the first isomorphism theorem.

3. Feb 5, 2013

### CAF123

Hi Kreizhn,

So applying $g_1 g_2^{-1}$ to the homomorphism, I get $$\theta(g_1 g_2^{-1}) = \theta(g_1) \theta(g_2^{-1}) = \theta(g_1) (\theta (g_2))^{-1} = h_1 (h_1)^{-1} = e_H,$$ where I defined $\theta(g_1) = \theta(g_2) = h_1$. Since we know only the identity in G, $e_G$ maps to $e_H$, I conclude that $g_1 = g_2$. Okay?

4. Feb 5, 2013

Looks good.

5. Feb 5, 2013

### CAF123

Thanks for the help.

6. Feb 14, 2013

### CAF123

Apparently I had to show that there existed a homormorphism between G and Imθ.

7. Feb 14, 2013

### Kreizhn

I don't follow: $\theta$ is your homomorphism, that is what you showed.

8. Feb 15, 2013

### CAF123

I don't think I showed $\theta$ was a homormorphism - it is given that $θ$ is a homormorphism. The comment on my work is that '$θ : G →H$ is a homormorphism, need to show homormorphism $G →\operatorname{Im} \theta$'

9. Feb 15, 2013

### jbunniii

There's not much to show. We can certainly define $\phi : G \rightarrow im(\theta)$ by $\phi(g) = \theta(g)$. Then $\phi$ is a homomorphism because $\phi(ab) = \theta(ab) = \theta(a)\theta(b) = \phi(a)\phi(b)$. The only distinction between $\phi$ and $\theta$ is that $\phi$ is surjective even if $\theta$ is not. Note that $\phi$ is injective if and only if $\theta$ is.

10. Feb 15, 2013

### CAF123

Could you elaborate further how you made these deducements?

11. Feb 15, 2013

### jbunniii

OK. I defined $\phi : G \rightarrow im(\theta)$ by $\phi(g) = \theta(g)$. If $a \in im(\theta)$ then by definition of the image, there is some $g \in G$ such that $\theta(g) = a$. But this means $\phi(g) = a$. This shows that $\phi$ is surjective.

Suppose $\theta$ is injective. If $\phi(g) = \phi(h)$ then $\theta(g) = \theta(h)$, so $g = h$ because of the injectivity of $\theta$. This shows that $\phi$ is injective.

Conversely, suppose $\phi$ is injective. If $\theta(g) = \theta(h)$, then $\phi(g) = \phi(h)$, and this implies $g = h$ because of the injectivity of $\phi$. This shows that $\theta$ is injective.

12. Feb 15, 2013

### CAF123

Makes sense, thanks. But why did you define $\phi(g) = \theta(g)$?Is this arbitrary?

13. Feb 15, 2013

### jbunniii

What other candidate is available if we want a homomorphism between $G$ and $im(\theta)$? We are not given any information about $G$ or $H$, other than the fact that $\theta : G \rightarrow H$ is a homomorphism. Therefore we have nothing to work with except $\theta$. There may well be other homomorphisms from $G$ onto $im(G)$, but no information is provided about any of them.

14. Feb 15, 2013

### Kreizhn

I think the gist of it is this:

Technically, when you restrict morphisms they could stop being morphisms. When you restrict the domain it may fail to be a subgroup so the map, while still defined, could fail to be a homomorphism. Conversely, if you restrict the codomain, the map may not have a well-defined target and so fail to be a mapping.

In this instance, we are restricting the codomain so that $\theta: G \to \text{im}\theta$. The only possible pitfall is that your target not be well-defined, for which it suffices to check that $\text{im}\theta \subseteq \text{im}\theta$. But this is a tautology, so there's nothing to show.

It seems incredibly pedantic, but perhaps if this is a "first course" in abstract algebra, the TA is just making sure you "dot your i's and cross your t's."

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