A result of a different nature is something called the Brauer trick, due (of course) to R. Brauer. It is a way to show that two (not necessarily normal) proper subgroups $A$ and $B$ of a finite group $G$ together generate a proper subgroup of $G$ using a character-theoretic argument.

The idea is to take a complex irreducible (non-trivial) character $\chi$ of $G$, and try show that $$\langle {\rm Res}_{A}^{G}(\chi),1\rangle + \langle {\rm Res}_{B}^{G}(\chi),1\rangle > \langle {\rm Res}_{A \cap B}^{G}(\chi),1\rangle .$$

If that inequality holds, then in the underlying $\mathbb{C}G$-module, $V$ say, associated with $\chi,$ we see $C_{V}(A)$ and $C_{V}(B)$ are both contained in $C_{V}(A \cap B),$ so by elementary linear algebra $$C_{V}(A) \cap C_{V}(B) \neq \{0\}.$$ In that case $\langle A,B \rangle$ fixes a non-zero vector $v$, so since $\chi$ is irreducible, the subgroup $\langle A,B \rangle$ is a proper subgroup of $G$.

Of course, here, there is no issue of whether $\langle A,B \rangle$ is a subgroup, but the character theory demonstrates that this subgroup is proper.

A result of a different nature is something called the Brauer trick, due (of course) to R. Brauer. It is a way to show that two (not necessarily normal) proper subgroups $A$ and $B$ of a finite group $G$ together generate a proper subgroup of $G,$ using a character-theoretic argument.

The idea is to take a complex irreducible (non-trivial) character $\chi$ of $G$, and try to show that $$\langle {\rm Res}_{A}^{G}(\chi),1\rangle + \langle {\rm Res}_{B}^{G}(\chi),1\rangle > \langle {\rm Res}_{A \cap B}^{G}(\chi),1\rangle .$$

If that inequality holds, then in the underlying $\mathbb{C}G$-module, $V$ say, associated with $\chi,$ we see $C_{V}(A)$ and $C_{V}(B)$ are both contained in $C_{V}(A \cap B),$ so by elementary linear algebra $$C_{V}(A) \cap C_{V}(B) \neq \{0\}.$$ In that case $\langle A,B \rangle$ fixes a non-zero vector $v$, so since $\chi$ is irreducible, the subgroup $\langle A,B \rangle$ is a proper subgroup of $G$.

Of course, here, there is no issue of whether $\langle A,B \rangle$ is a subgroup, but (when applicable) the character theory demonstrates that this subgroup is proper.

A result of a different nature is something called the "Brauer trick", due (of course) to R. Brauer. It is a way to show that two (not necessarily normal) proper subgroups $A$ and $B$ of a finite group $G$ together generate a proper subgroup of $G$ using a character-theoretic argument.

The idea is to take a complex irreducible (non-trivial) character $\chi$ of $G$, and try show that $$\langle {\rm Res}_{A}^{G}(\chi),1\rangle + \langle {\rm Res}_{B}^{G}(\chi),1\rangle > \langle {\rm Res}_{A \cap B}^{G}(\chi),1\rangle .$$

If that inequality holds, then in the underlying $\mathbb{C}G$-module, $V$ say, associated to $\chi,$ we see $C_{V}(A)$ and $C_{V}(B)$ are both contained in $C_{V}(A \cap B),$ so by elementary linear algebra $C_{V}(A) \cap C_{V}(B) \neq \{0\}.$ In that case $\langle A,B \rangle$ fixes a non-zero vector $v$, so since $\chi$ is irreducible, the subgroup $\langle A,B \rangle$ is a proper subgroup of $G$.

Of course, here, there is no issue of whether $\langle A,B \rangle$ is a subgroup, but the character theory demonstrates that this subgroup is proper.

A result of a different nature is something called the Brauer trick, due (of course) to R. Brauer. It is a way to show that two (not necessarily normal) proper subgroups $A$ and $B$ of a finite group $G$ together generate a proper subgroup of $G$ using a character-theoretic argument.

The idea is to take a complex irreducible (non-trivial) character $\chi$ of $G$, and try show that $$\langle {\rm Res}_{A}^{G}(\chi),1\rangle + \langle {\rm Res}_{B}^{G}(\chi),1\rangle > \langle {\rm Res}_{A \cap B}^{G}(\chi),1\rangle .$$

If that inequality holds, then in the underlying $\mathbb{C}G$-module, $V$ say, associated with $\chi,$ we see $C_{V}(A)$ and $C_{V}(B)$ are both contained in $C_{V}(A \cap B),$ so by elementary linear algebra $$C_{V}(A) \cap C_{V}(B) \neq \{0\}.$$ In that case $\langle A,B \rangle$ fixes a non-zero vector $v$, so since $\chi$ is irreducible, the subgroup $\langle A,B \rangle$ is a proper subgroup of $G$.

Of course, here, there is no issue of whether $\langle A,B \rangle$ is a subgroup, but the character theory demonstrates that this subgroup is proper.