Example of Homomorphisms Studying homomoprhisms, and there are may examples of them given in the textbook, but these two examples in particular I am not sure if they are homomorphisms. I would say $S_5 $ example is a homomorphism, whereas $GL(n,R)$ example is not a  homomorphism. 
$\Phi:GL(n,R) \rightarrow \mathbb{R^*}$ for $\Phi(A) = Det(A)^{10}$
$\Phi:S_5 \rightarrow {S_5}$ for $\Phi(\sigma) = \sigma^{120}$
My reason is, for the symmetric group $S_5$, when you take the power of an element from the symmetric group $S_5$, you will eventually get some member in the symmetric group in $S_5$. However, the general linear group is not a homomorphism for $A$ and $B$ in $GL(n,R)$; as $\Phi(AB) \ne Det(A)^{10}Det(B)^{10}$, Is this correct?
 A: $\Phi: S_5 \to S_5;\, \Phi(\sigma) = \sigma^{120}$ is secretly the identity map, since $g^{|G|} = 1$ for any group $G$ and group element $g \in G$. For this reason, it's a homomorphism. 
In general, though, the "$n$th power map" $\Phi: G \to G;\, \Phi(g) = g^n$ need not be a homomorphism for nonabelian groups $G$, since it's often not the case that $(ab)^n = a^nb^n$ (while such a thing is guaranteed in an abelian group). 
Try the map $\Phi(g) = g^2$ for the dihedral group of order $8$. If $a$ is a 1/4-turn rotation and $b$ any reflection, then $ab$ is a reflection, hence $\Phi(ab) = (ab)^2 = 1$, while $\Phi(a)\Phi(b) = a^2 b^2 = a^2$ is a 1/2-turn rotation, and not equal to $\Phi(ab)$.

Considering $\Phi : GL(n, R) \to GL(n, R);\, \Phi(A) = \det(A)^{10}$, let's check:
$$\Phi(AB) = \det(AB)^{10} = (\det(A)\det(B))^{10} = \det(A)^{10}\det(B)^{10} = \Phi(A)\Phi(B).$$
The secret sauce here is really that $\Phi$ is a composition of two maps: The determinant map, and the $n$th power map. Since the determinant map is a homomorphism that takes us to an abelian group, the $n$th power map that ensues is a homomorphism. Hence $\Phi$ is a composition of homomorphisms, and is itself a homomorphism.
