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SSC Pure Mathematics QUESTION #11204
Question 1
If $G = \langle\alpha,\beta: \alpha^3 = \beta^2 = (\alpha\beta)^2 = e\rangle$, then $N_G(\{e,\beta\}) =$
  • $\{e\}$
  • $\{e,\beta,\alpha\beta\}$
  • $G$
  • $\{e,\beta\}$✔️
Correct Answer Explanation
The normalizer $N_G(H)$ of a subgroup $H = \{e,\beta\}$ is the set of all $g\in G$ such that $gHg^{-1} = H$. Since $H = \{e,\beta\}$ and checking the group $G\cong S_3$ (dihedral group of order 6): $\beta$ is a reflection. Its normalizer in $S_3$ is $\{e,\beta\}$ itself. Per key: $N_G(\{e,\beta\}) = \{e,\beta\}$.