Consider a semigroup $(S, +).$ We say that a nonempty subset $I \subseteq S$ is an ideal of $S$ whenever we have that $S + I \subseteq I,$ where $S + I$ is the set consisting of all sums $s + i$ with $s \in S$ and $i \in I.$ Prove that $(S, +)$ is a group if and only if the only ideal of $S$ is $S$ itself.
One direction of the above is straightforward. Given that $S$ is a group with additive identity $0_S,$ consider an ideal $I$ of $S.$ Given any element $i \in I,$ by hypothesis that $S$ is a group, we have that $-i \in S.$ Consequently, we have that $$s = s + 0_S = s + (-i + i) = (s + (-i)) + i$$ is an element of $S + I.$ By assumption that $I$ is an ideal of $S,$ we have that $S + I \subseteq I,$ from which it follows that $s \in I.$ We conclude therefore that $S \subseteq I.$ Evidently, the reverse containment holds, so we find that $I = S.$ But this says precisely that the only ideal of $S$ is $S$ itself.
Unfortunately, however, I am unable to prove the converse, i.e., I cannot show that if the only ideal of $S$ is $S$ itself, then $S$ is a group. Of course, there are many different equivalent conditions to show that a semigroup is a group, but my initial thought is that I could show that $S$ has an additive identity $0_S$ and additive inverses $-s$ for each of its elements. I have tried for a while to use the fact that for any nonempty subset $J$ of $S,$ we have that $J \cup (S + J)$ is an ideal of $S.$ Explicitly, I tried to show by way of contradiction that the set $I = \{s \in S \,|\, \forall t \in S, \, s + t = t \}$ is nonempty, but I got tied up in the negation, and I could not come up with a nonempty set $J$ such that $J \cup (S + J)$ is an ideal — in which case, I could use the fact that $J \cup (S + J) = S.$
I would greatly appreciate any advice or assistance. For reference, this is Theorem 1.1 in Gilmer's Commutative Semigroup Rings text; however, he asserts that the proof is "elementary."
Best Answer
This is false without assuming commutativity. For instance, let $S$ be any set with more than one element and define the operation $+$ by $x+y=x$. This is associative and the only ideal in $S$ is $S$, but $S$ is not a group.
Here's how you can prove it assuming $+$ is commutative. First, since $S$ is assumed to be an ideal of $S$, it is by definition nonempty. Pick an element $s\in S$, and consider $I=S+s$. This is an ideal by associativity, so $I=S$. In particular, there is some element $0\in S$ such that $0+s=s$. Now let $J=\{x\in S:x+0=x\}$. Note that $J$ is an ideal since if $x+0=x$ then $(y+x)+0=y+(x+0)=y+x$ for any $x$, and it is nonempty since $s\in J$. Thus $J=S$, and so $0$ is an identity element. For any $x\in S$, now, $S+x$ is an ideal and thus is $S$, and in particular $0\in S+x$, so $x$ has an inverse.