If \(H<G\) and \([G:H]=2\), show that \(H\triangleleft G\).
Proof: If \([G:H]=2\), then \(gH=Hg\) for all \(g\in G\), so \(H\triangleleft G\).

【Basic Abstract Algebra】Exercises for Section 3.1 — Cosets and Lagrange's Theorem - 只會加減乘除 - 博客園 (cnblogs.com)
If \([G:H]=2\), then \(gH=Hg\) for all \(g\in G\).

Find out all normal subgroup of \(A_4\), and give all factor groups of \(A_4\) over its normal group.

Solution: \(A_4=\{(1),(123),(132),(124),(142),(134),(143),(234),(243),(12)(34),(13)(24),(14)(23)\}\), and \(|A_4|=4!=24\). By the Lagrange's Theorem, the possible order of subgroup of \(A_4\) are \(1,2,3,4,6,12\).
Find subgroups:

• order \(1\): \(\{(1)\}\);
• order \(2\): \(\{(1),(12)(34)\},~\{(1),(13)(24)\},~\{(1),(14)(23)\}\);
• order \(3\): \(\{(1),(123),(132)\},~\{(1),(124),(142)\},~\{(1),(134),(143)\},~\{(1),(234),(243)\}\);
• order \(4\): \(\{(1),(12)(34),(13)(24),(14)(23)\}\);
• order \(6\): Does not exist. Assume \(H<A_4\) and \(|H|=6\), then \([A_4:H]=2\). We will show that \((123)\in H\). Since \([A_4:H]=2\), we have \(A_4=H\cup gH\) for some \(g\in A_4\) and \(H\triangleleft A_4\). If \((123)\notin H\), then \((132)=(123)^{-1}\notin H\) and \(A_4=H\cup (123)H\) & \(H\cap(123)H=\varnothing\). Then \(\exists h\in H\), s.t. \((132)=(123)h\Rightarrow h=(123)\), it is contradict to \((123)\notin H\). Thus \((123)\in H\). Similarly, \((123),(132),(124),(142),(134),(143),(234),(243)\in H\), then \(|H|\ge 9\), \(\to\leftarrow\).
• order \(12\): \(A_4\).

Find normal subgroups:

• order \(1\): \(\{(1)\}\);
  • factor group: \(A_4/\{(1)\}\cong A_4\).
• order \(4\): \(V_4=\{(1),(12)(34),(13)(24),(14)(23)\}\);
  • factor group: \(A_4/V_4=\{V_4,(123)V_4,(132)V_4\}\).
• order \(12\): \(A_4\).
  • factor group: \(A_4/A_4=\{A_4\}\).