4 part problem regarding homomorphisms

[AutGuy98]

Hey guys,

I have some more problems that I need help with figuring out what to do. The first one is divided into 4 mini-problems, sub-sections, whatever you would like to call them. It asks:

(a) Show that (Z/4Z,+) is not isomorphic to ((Z/2Z) x (Z/2Z),+). Find a homomorphism from (Z/4Z,+) to ((Z/2Z) x (Z/2Z),+).
(b) Let G and H be groups and let $\phi:G\to H$ be a homomorphism. Show that $ker(\phi)$ is a normal subgroup of G and that $\phi(G)$ is a subgroup of H.
(c) Find a pair of groups G,H and a homomorphism $\phi:G\to H$ between them such that $\phi(G)$ is not normal in H.
(d) Let G and G' (i.e. G prime) be finite groups whose orders have no common factor. Prove that the only homomorphism $\phi:G\implies G'$ is the trivial one (i.e. $\phi(x)=1$ for all x's in G).

I would greatly appreciate it if someone could please get back to me about these by tomorrow. But if more time is needed to work the problems out, I completely understand. Thank you in advance to whomever assists me with these. I am extremely grateful.

 

[jvicens]

I am happy to assist you with these problems. Let's go through them one by one.

(a) To show that (Z/4Z,+) is not isomorphic to ((Z/2Z) x (Z/2Z),+), we need to find a property that is true for one group but not the other. In this case, it is the order of the elements. In (Z/4Z,+), the element 2 has an order of 4, meaning that when we add it to itself 4 times, we get the identity element. However, in ((Z/2Z) x (Z/2Z),+), all elements have an order of either 1 or 2. Therefore, these two groups cannot be isomorphic.

To find a homomorphism from (Z/4Z,+) to ((Z/2Z) x (Z/2Z),+), we can define $\phi:Z/4Z \to (Z/2Z) x (Z/2Z)$ as $\phi([x]) = ([x],[x])$. This is a valid homomorphism because for any $[x],[y] \in Z/4Z$, we have $\phi([x]+[y]) = ([x+y],[x+y]) = ([x],[x]) + ([y],[y]) = \phi([x]) + \phi([y])$. You can verify that this is indeed a homomorphism.

(b) To show that $ker(\phi)$ is a normal subgroup of G, we need to show that for any $g \in G$ and $k \in ker(\phi)$, we have $gkg^{-1} \in ker(\phi)$. This is true because $\phi(gkg^{-1}) = \phi(g)\phi(k)\phi(g)^{-1} = \phi(g)1\phi(g)^{-1} = \phi(g)\phi(g)^{-1} = 1$, since $\phi(k) = 1$ by definition of $ker(\phi)$.

To show that $\phi(G)$ is a subgroup of H, we need to show that it is closed under the group operation and contains the identity element. Both of these are true since $\phi$ is a homomorphism and $\phi(1_G) = 1_H$.

(c) A simple example of a