it happens for a wide class of groups. there is always a homomorphism N-->Aut(H) for any two groups N,H, but often this is only the trivial one:

\(\displaystyle \phi(n) = \text{id}_H\) for all \(\displaystyle n \in N\). suppose we have p = 2.

then p^3 = 8, so we're looking for a nontrivial homomorphism of Z2 into Aut(Z4). now φ(4) = 2, so we have just 2 automorphisms:

x-->x (this is x--->(1+0*2)x)

x-->3x (this is x--->(1+1*2)x).

note that x-->3x is really the same as x--> -x in Z4.

so we can re-write our multiplication in Z4 x Z2 as: (a,b)*(a',b') = (a+((-1)^b)(a'), b+b').

we have two types of elements: (c,0) and (c,1).

multiplying by (c,0) just "collects first coordinates": (c,0)*(a,b) = (c+a,b).

multiplying by (c,1) "twists the first coordinate, and flips the second": (c,1)*(a,b) = (c-a,b+1).

this is our old friend D4, with (1,0) being the rotation r, and (0,1) being the reflection s.

in **drexel28's** post, we want to have \(\displaystyle \ell|\varphi(k)\) or else there may not be a non-trivial homomorphism,

since \(\displaystyle \varphi(\mathbb{Z}_{\ell})\) is a subgroup of Aut(Zk) and so has to divide its order φ(k). if indeed \(\displaystyle \ell|\varphi(k)\), then we can just send [1] to an element of an order that divides \(\displaystyle \ell\).