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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:

The problem reads as follows:

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.

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    $\begingroup$ 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. $\endgroup$ – Santana Afton Oct 23 '19 at 14:47
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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.

In general, the difference between the conjunction and biconditional is the conjunction is true only when both parts are true whereas the biconditional statement is true whenever both parts have the same truth value.

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