# Normal subgroup $N$ of a p-group $G$ intersects $Z(G)$ nontrivially; What is wrong with the following trivial argument?

I'm trying to show that a normal subgroup $$N$$ of a p-group $$G$$ intersects $$Z(G)$$ nontrivially (please don't tell how to show it), but it seem it is quite a trivial question considering the following argument:

Consider $$Z(N)$$, a nontrivial subgroup of $$N$$. Since any element in $$Z(G)$$ also commutes with the elements of $$N$$, $$Z(G) \subseteq Z(N)$$, but we know that $$Z(N) \leq N$$, so $$\{e\} \not = Z(G) \subseteq N.$$

If the answer was this, I don't think my Algebra professor would ask it, so what is wrong with the above argument ?

• It's certainly not the case that $Z(G)\subseteq Z(N)$. – Lord Shark the Unknown Dec 15 '18 at 6:08
• @LordSharktheUnknown Why ? – onurcanbektas Dec 15 '18 at 6:08
• @LordSharktheUnknown By definition, any element that commutes with all the elements in the group shouldn't commute with the elements in the subgroup ? – onurcanbektas Dec 15 '18 at 6:09
• What if $N$ is a proper subgroup of $Z(G)$? – Lord Shark the Unknown Dec 15 '18 at 6:13
• An element $g$ of $G$ belongs to $Z(N)$ if and only if it commutes with all the elements of $N$ and it belongs to $N$. If the latter condition is not verified you only know that $g \in C_G(N)$ which is in general bigger than $Z(N)$. – Pietro Gheri Dec 15 '18 at 12:40

$$Z(N)$$ and $$Z(G)$$ need not be at all related. If $$N$$ is abelian, then $$Z(N) =N$$, but the center of $$G$$ might intersect $$N$$ trivially. For example, the center of $$S_3\times \mathbb Z_3$$ intersects a subgroup of order $$2$$ in the first factor trivially.
In general, $$Z(G)$$ is a subgroup of the centralizer of $$N$$, but not necessarily the center.
• In my notation, $Z(G):= C(G)$. – onurcanbektas Dec 15 '18 at 14:49
• @onur Yes, but $Z(N) \neq C(N)$ in general, unless you're using highly unusual notation. – Matt Samuel Dec 15 '18 at 14:51
• what is $Z(N)$ in your notation ? – onurcanbektas Dec 15 '18 at 14:53
• @onur The set of all elements of $N$ that commute with every element of $N$. $C(N)$ is the set of all elements of $G$ that commute with every element of $N$. – Matt Samuel Dec 15 '18 at 14:56