# Understanding the biconditional logical connective in the set: $K = \{x \in G: x \circ a \circ x^{-1} \in H \iff a \in H \}$

There is a problem in Pinter's A Book of Abstract Algebra (Chpt 5 Problem D7) that asks to prove that a set is subgroup. My question is not about the specific objective of the exercise but rather about the notation used in the exercise:

Let $$H$$ be a subgroup of $$G$$, and let $$K = \{x \in G: x \circ a \circ x^{-1} \in H \iff a \in H \}$$

The exercise wants you to subsequently demonstrate that $$K$$ is a subgroup of $$G$$ and $$H$$ is a subgroup of $$K$$.

I hope this does not come off as an "opinion question", but why did the author choose to write the such that statement of the set in the form of a biconditional:

$$x \circ a \circ x^{-1} \in H \iff a \in H$$

Couldn't Pinter have just as easily used a conjunction, $$\land$$ , symbol? Would the meaning of the question have changed if the such that statement read as:

$$x \circ a \circ x^{-1} \in H \land a \in H$$

Is there some sort of nuance that I am missing...because it seems as though the proof would proceed unchanged. I get that the truth table values for $$\land$$ and $$\iff$$ are not the same...but I can't see how this affects the proof itself.

I have never previously seen a logical connective in the S.T. statement of a set in the form of a biconditional...and I do not really understand why it is there.

• You might be interested in this question regarding the same exercise. In the first edition, the ‘if and and only if’ was originally a ‘for every.’ It seems that the biconditional is preferred; see a counterexample regarding the ‘for every’ statement here. – Santana Afton Oct 23 '19 at 14:47

Firstly, the biconditional statement $$xax^{-1}\in H\iff a\in H$$ is apparently meant to be read as $$\mathbf B(x,a):(\forall a\in G \;\; xax^{-1}\in H\iff a\in H)$$ with $$\forall$$ unnecessary as $$G$$ is the implicit universe. Replacing $$\mathbf B(x,a)$$ with $$\mathbf C(x,a) : (xax^{-1}\in H\land a\in H)$$ in $$K$$ we obtain a $$\it different$$ set, $$K':=$$ {$$x\in G : \forall a\in G\mathbf \; C(x,a)$$} $$=\emptyset$$ or $$G$$, depending on whether $$H$$ is a proper subgroup or not respectively. To see this note that when $$H$$ is proper there is of course no way $$\it every$$ member of $$G$$ is simultaneously a member of $$H$$. When $$H=G$$ then $$\mathbf C(x,a)$$ is true for all $$x,a$$ and we have $$K'=G$$ with of course $$H\leq K'$$.
Finally, to see how this affects the proof note that when $$H$$ is a proper subgroup and we replace $$\mathbf B(x,a)$$ with $$\mathbf C(x,a)$$ in $$K$$ then there is no proof because the empty set is not a group and has no subgroups.