Is there an corresponding concept of prime ideals in groups? Let $G$ be a group and $R$ a commutative ring.
$G$ is simple if it has no non-trivial proper subgroups. It can be shown that $R$ is a field if it has no non-trivial proper ideals. Furthermore, if $N \unlhd G$ is maximal, $G/N$ is simple and if $I \unlhd R$ is  maximal, $R/I$ is a field, in both cases by the appropriate correspondence theorem.
This seems to suggest a sort of correspondence between fields and simple groups.
We also have the result that if $I \unlhd R$ is prime, $R/I$ is an integral domain.
Question: Can prime ideals or integral domains in rings also be identified with any particular structures in groups?
Since ideals are generally considered to be the corresponding concept in rings for normal subgroups, this might suggest that prime ideals correspond to normal subgroups with a further condition, but one weaker than maximality.

I am aware there is a similar question already on MSE, but it did not ask about prime ideals.
 A: As you note, yhe characterizations you provide are consequences, rather than the definitions, of the notions. Moreover, the one for prime ideals only holds when $R$ is commutative with unity. For example, in the ring of $2\times 2$ matrices with real coefficients, the zero ideal is maximal, but the ring is not field or even a ring with no zero divisors.

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*A maximal ideal $M$ of a ring $R$ is an ideal such that $M\neq R$, and if $I$ is any ideal with $M\leq I\leq R$, then either $M=I$ or $I=R$. The corresponding notion for groups is maximal normal subgroup (which is a normal subgroup $N$ such that $N\neq G$ and if $N\leq M\leq G$ with $M\triangleleft G$, then either $N=M$ or $M=G$); this is equivalent to $G/N$ being simple, just as in the ring case $M$ being maximal is equivalent to $R/M$ being simple (it is only a field when $R$ is commutative with unity).


*A prime ideal $P$ of a ring $R$ is an ideal such that $P\neq R$, and whenever $I$ and $J$ are ideals of $R$, if $IJ\leq P$, then either $I\leq P$ or $J\leq P$. The corresponding notion for groups would be that a normal subgroup $N$ of $G$ is "prime" if and only if whenever normal subgroups $M_1$ and $M_2$ are such that $M_1M_2\leq P$, we have either $M_1\leq P$ or $M_2\leq P$. As Jonas Linssen notes, this holds for every subgroup, since $M_i\leq M_1M_2$ (which is not true for ideals in rings).


*You have the stronger notion of completely prime ideal: an ideal $P$ is completely prime if $P\neq R$ and for every $a,b\in R$, if $ab\in P$ then $a\in P$ or $b\in P$ (completely prime ideals are necessarily prime ideals; in the commutative case the two notions are equivalent, but not in the noncommutative case: in the ring of $2\times 2$ matrices over a field, the zero ideal is prime but not completely prime). So you could ask for a normal subgroup $N\neq G$ such that whenever $xy\in N$, either $x\in N$ or $y\in N$. But now this holds for no proper normal subgroup, since you can take any $x\notin N$, and then $xx^{-1}\in N$ but neither $x$ nor $x^{-1}$ are in $N$.


*You could weaken the condition from commutative rings in which we ask that $R/P$ have no nonzero zero divisors to asking that it have no nonzero nilpotent elements. That would be equivalent to asking that $G/N$ have no elements of finite order (that is, be torsionfree). This is a relevant notion: it requires that $N$ contain every torsion element of $G$, and be "isolated": a subgroup $H$ of $G$ is isolated if and only if for every $a\in G$, either $a\in H$ or $\langle a\rangle\cap H = \{e\}$. If a normal subgroup $N$ contains every torsion element and is isolated, and $g\notin N$, then $g$ is of infinite order and $\langle g\rangle\cap N=\{e\}$, so $gN$ has infinite order in $G/N$. Conversely, if $g\notin N$ has finite order, then $gN$ is nontrivial and torsion in $G/N$; and if $N$ is not isolated, then letting $a\notin N$ with $e\neq a^k\in N$, then $aN$ is nontrivial but has finite order in $G/N$, so $G/N$ is not torsionfree in either case.
I will note that the multiplicative structure of a ring is of course a semigroup (or a monoid if you required identities), and there is a notion of "prime ideal" in semigroups.
In a semigroup $S$, a nonempty subset $A$ is a "left ideal" if $SA\subseteq A$ (where $SA=\{sa\mid s\in S, a\in A\}$; a "right ideal" if $AS\subseteq A$, and a "two-sided ideal" (or simply "ideal") if it is both a left ideal and a right ideal.
A left/right/two-sided ideal $P$ of a semigroup $S$ is prime if whenever $ab\in P$, either $a\in P$ or $b\in P$; that is, precisely the analogy of "completely prime ideal" for rings.  And alternative definition is similar to the one for prime ideal, that $P$ is prime if whenever $AB\subseteq P$ for ideals $A$ and $B$ of $S$, either $A\subseteq P$ or $B\subseteq P$. These do play an interesting role in semigroups.
