(1) Prove that if Z(G)

1
Before proceeding to answer the first question let's look at a lemma.
Lemma .1 Let p be a prime and G be a nonabelian group of order p3 with center Z(G). Then |Z(G)| = p,
G/Z ∼= Zp × Zp.
Proof Since |G| is of the form pn therefore by class equation we have |Z(G) 6= {e}. Therefore we have
|Z(G)| = p, p2, or p3. Since G is non-abelian therefore |Z(G)| 6= p3. And if |Z(G)| = p2 then |G/Z(G)| = p
which is cyclic =⇒ G is abelian, hence |Z(G)| 6= p2. Therefore the only choice for |Z(G)| = p. Now the group
G/Z(G) has order p2 and we know that any group of order p2 is isomorphic to Zp2 or Zp × Zp. Since G/Z(G)
is non-cyclic therefore G/Z(G) ∼= Zp × Zp.
Now for the answer of your question.
1. Consider a non-abelian group of order 27. By above lemma |Z(G)| = 3. And we have
|G/Z(G)| = 9 ∼= Z3 × Z3.
2 (a). By definition we have Z
(
G/N
)
=
{
x ∈ G/N | xy = yx ∀ y ∈ G
}
, So let Ng ∈ Z
(
G/N
)
. Then we have
Ngy = yNg
[
for all y ∈ G
]
=⇒ N(gyg−1) = yN
=⇒ N(gyg−1) = Ny
=⇒ N(gyg−1y−1) = N
=⇒ gyg−1y−1 ∈ N
This says that [g, y] ∈ N for all y ∈ N.
2 (b). Consider a...