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It is well known that if $G$ is a finite p-group and $1 \neq H \unlhd G$ then $H \cap Z(G) \neq 1$.

Are there other families of groups with this property?

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(All groups are finite.) An alternative characterization is that the socle is central. For instance, any quasi-simple but not simple group has this property.

The socle is the subgroup generated by the minimal normal subgroups. If a minimal normal subgroup intersects the center, it does so in a normal subgroup, and hence is in fact contained in the center. In a group in which every non-identity normal subgroup intersects the center non-trivially, every minimal normal subgroup is contained in the center, and thus the socle too must be contained in the center. In the other direction, if the socle is contained in the center, then every normal subgroup contains a minimal normal subgroup, and thus intersects the center non-trivially.

An explicit example is the group SL(2,5) of order 120. It is far from being a p-group, as it is a perfect group. However, its normal structure is very restricted. It is “sitting on top of its center”, so that its center is literally its socle.

In general, any perfect group G such that G/Z(G) is simple but Z(G)≠1 is also an example of a simple group sitting proudly atop its central socle. These are called quasi-simple groups.

For perfect groups, you just need a proper “stem” extension of a perfect group. That is, take any centerless perfect group with non-trivial Schur multiplier, and take a (even partial) covering group. The original perfect group did not have to be simple.

The same will be true for any group that is a direct product of its Sylow subgroups (apply the result on $p$-groups to the intersections of the normal subgroup with the Sylows). This applies e.g. to nilpotent groups, as well as to hypercentral groups.

~~Further, if $G$ has this property and $S$ is simple, then $G\times S$ also has this property: indeed the centre of $G\times S$ is $Z(G)\times\{1\}$. Also, any normal subgroup of $G\times S$ has trivial projection modulo $G$, so is contained in $G$.~~ See Jack’s comment.

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