a question about conjunction logic Let's take this question as an example:

prove that a finite group is the union of proper subgroups if and only if the group is not cyclic.

Then we prove that If

*

*G is the union of proper subgroups

*G is simutaneously a cyclic group

Then we would have a contradiction, in other words, G cannot be simutaneously both a union of proper subgroups and cyclic.
What I want to ask is, after this contradiction, aren't we have already proved the pervious statement? since If G is indeed a union of proper subgroups, then G must not by cyclic, and if G is indeed cyclic, then G must not be a union of proper subgroups.
Or, that this contradiction  only proved this argument only in one direction, and says nothing at all in another direction?
 A: Let G be a finite group.
Let A be the proposition that $A :=$ G is a union of proper subgroups. 
Let B be the proposition that $B :=$ G is not cyclic.
You want to prove that $A \Longleftrightarrow B $
What you are doing is to prove $A \land \lnot B \implies \bot$, that is equivalent to proving $\lnot (A \land \lnot B)$, which is equivalent to proving  $\lnot A \lor B$ which is equivalent to proving  $A \implies B$ .
So you're proving only one direction of the statement. You also need to prove  $B \implies A$, that is the contradiction $B \land \lnot A \implies \bot$.
A: Your argument would show that no group can be both cyclic and also the union of proper subgroups. That's not enough, however, as the problem asks you to prove that any group has exactly one of those two properties, not just that no group has both properties.
I would instead do it by proving the following two statements:

*

*A cyclic group is not the union of proper subgroups. Specifically, there are elements in the group such that no proper subgroup contains those elements. (Hint: what are the "special" elements that cyclic groups have that other groups don't?)

*A non-cyclic group is the union of proper subgroups. Specifically, in a non-cyclic group, each and every element lies in at least one proper subgroup, and you can take the union of those. (Hint: a cyclic subgroup must be proper.)

