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 $$Z(G)=\{a\in G\mid \forall b\in G, ab=ba\}.$$


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 $$\overline{Z(G)}\le Z(\overline{G}).$$


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.

Please help :)

  • 1
    $\begingroup$ I think it is easier to prove it directly i.e. let $gK\in Z(G)K/K$ and prove that $gK \in Z(G/K)$. $\endgroup$
    – Derek Holt
    Oct 25, 2020 at 16:31

1 Answer 1


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 $G$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .