2.29 (1966)

No. Let $B$ be a finite group such that $B = [B, B] \neq 1$ and let $r$ be the rank of $B$. Put $A = \bigoplus_{p \mid |B|} (\mathbb{Z}/p\mathbb{Z})^{r+1}$ and $G = A \wr B$. We define a homomorphism $\varphi : G \to A$ by setting $(bf)^\varphi = \sum_{x \in B} f(x)$, where $b \in B$ and $f \in F = \text{Fun}(B, A)$. Suppose that $f^b = f$ for $b \in B$ and $f \in F$. It is clear that $f^\varphi \in nA$, where $n = |b|$. In particular, $C_F(b)^\varphi \leqslant pA \leqslant O_{p'}(A)$ if $p \mid n$. Every proper abelian subgroup $H$ of $G$ is contained in a proper normal subgroup. Indeed, we may assume that $H \not\leqslant F$. We fix an element $bf \in H$, where $f \in F$, $b \in B$, $b \neq 1$. Let $p$ be a prime dividing $|b|$. For any $h \in H \cap F$ we have $bfh = hbf = bhbf = bfh^b$, whence $h = h^b$. Hence $(H \cap F)^\varphi \leqslant C_F(b)^\varphi \leqslant O_{p'}(A)$. If $T$ is the full preimage of $O_{p'}(A)$ in $G$, then $G/T \cong (\mathbb{Z}/p\mathbb{Z})^{r+1}$ and therefore $H/H \cap T$ is an elementary abelian $p$-group. The rank of it is $\leqslant r$, since $H/H \cap T$ embeds into $B$ and $H \cap F \leqslant H \cap T$. Thus, $HT$ is a proper normal subgroup containing $H$. On the other hand, $F$ is a proper abelian subgroup that is not contained in any proper normal subgroup of prime index, since $G/F \cong B = [B, B]$. (G. M. Bergman, I. M. Isaacs, Letter of June, 17, 1974.)