# Let $K\unlhd G$. Show $Z(G)K/K\le Z(G/K)$. (The Second Isomorphism Theorem, perhaps?)

This is Exercise 150 of Rose's "A Course on Group Theory." According to Approach0, it is new to MSE.

## The Details:

Since terminology differs, the Second Isomorphism Theorem, as stated on page 56 ibid., is:

Theorem: Let $$H\le G$$ and $$K\unlhd G$$. Then $$H\cap K\unlhd H$$ and $$H/H\cap K\cong HK/K$$.

For clarity:

Definition: We denote by $$Z(G)$$ the centre of $$G$$, given by $$Z(G)=\{a\in G\mid \forall b\in G, ab=ba\}.$$

Also:

Definition 2: A subgroup $$H$$ of a group $$G$$ is normal in $$G$$, written $$H\unlhd G$$, if $$aH=Ha$$ for all $$a\in G$$.

## The Question:

Suppose that $$K\unlhd G$$. Let $$\overline{G}=G/K$$ and $$\overline{Z(G)}=Z(G)K/K$$. Show that $$\overline{Z(G)}\le Z(\overline{G}).$$

## Thoughts:

This exercise appears directly below the statement & proof of the Second Isomorphism Theorem; that, together with its apparent suitability for the Theorem, makes me think that it is relevant to the exercise.

So, by the Second Isomorphism Theorem, we have, since $$K, Z(G)\unlhd G$$, that

$$\overline{Z(G)}=Z(G)K/K\cong Z(G)/Z(G)\cap K.$$

It is not clear to me where to go from here.

That said, since $$Z(G)\subseteq G$$ and $$Z(G)\cap K\subseteq K$$, (I think) we can deduce that

$$Z(G)/Z(G)\cap K\subseteq G/K=\overline{G}.$$

I don't think this tells us much.

Another thought I have is to try & construct an isomorphism explicitly, rather than relying too heavily on the isomorphism theorem(s). This isomorphism might be a composition of the homomorphism used to prove the Second Isomorphism Theorem with some other, handy homo-/isomorphism. Other than that, I don't see how to develop the idea further. This idea is somewhat unclear, even to me.

My go-to technique for proving $$\mathcal{H}\le \mathcal{G}$$ as groups is the one-step subgroup test.

• I think it is easier to prove it directly i.e. let $gK\in Z(G)K/K$ and prove that $gK \in Z(G/K)$. Oct 25, 2020 at 16:31
Let $$H = Z(G) \cap K$$, you already have that $$Z(G)/H \subseteq G/K$$ then, $$\forall gH, kH \in Z(G)/H$$ we have, $$k^{-1}H\ast gH = k^{-1}gH$$ (as $$k \in Z(G) \Rightarrow k^{-1} \in Z(G)$$) but $$k^{-1}gH$$ is still in $$Z(G)/H$$ as $$k^{-1}g$$ is also in the center of $$G$$.