Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 3 / 8D (Groups)

When $[G:H]=2$, there are exactly two left cosets and two right cosets. For $g\notin H$, both $gH$ and $Hg$ are the complement $G\setminus H$; for $g\in H$, both equal $H$. This is the index-two subgroup is normal argument: $gH=Hg$ for every $g$, which is equivalent to $gH{g}^{-1}=H$. Therefore every subgroup of index two is normal.

In the dihedral group $D_{2n}=⟨r,s⟩$, the rotation cyclic subgroup $⟨r⟩$ has order $n$ and index two, so it is a normal subgroup. It is nontrivial and proper for the usual dihedral convention $n\ge 2$ (polygon symmetries have $n\ge 3$). If the degenerate convention $D_2=C_2$ were allowed, the claimed nontrivial proper subgroup would not exist; the rotation argument uses $n>1$.

For any $k\ge 3$, let $G=S_k$ and take the stabilizer subgroup $H$ of the letter $1$. Its elements freely permute the other $k-1$ letters, soConjugation by $(12)$ sends $H$ to the stabilizer of $2$, which is different: for instance $(23)$ belongs to $H$ but does not fix $2$. Thusgives a nonnormal subgroup of every prescribed index $k\ge 3$.

A simple group has no normal subgroup except the identity subgroup and the whole group. To prove the Simplicity of the alternating group A5, we compute its conjugacy classes. Since the sign homomorphism is onto, $|A_5|=5!/2=60$. Its possible nonidentity cycle types are a three-cycle, a product of two disjoint transpositions, and a five-cycle.

There are $\left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 2=20$ three-cycles. The centralizer in $S_5$ of $(123)$ consists of its three powers and an independent interchange of the two unused letters. Only the three powers are even. Its centralizer in $A_5$ therefore has order three, giving one conjugacy class of size $60/3=20$ by the orbit-stabilizer theorem.

There are $5\cdot 3=15$ products of two disjoint transpositions: choose the fixed letter and then a pairing of the other four letters. For $(12)(34)$, a commuting permutation fixes $5$ and may interchange the two pairs and swap within them, giving a centralizer of order eight in $S_5$. It contains an odd element, $(12)$, so exactly half its elements are even. The centralizer in $A_5$ has order four, and these fifteen elements form one conjugacy class of size $60/4=15$.

Finally, there are $4!=24$ five-cycles. A permutation commuting with a five-cycle is determined by the image of one letter and must be a power of that cycle. All five powers are even, so the centralizer in $A_5$ has order five. Each such conjugacy class has size $60/5=12$, giving two classes of size twelve.

A normal subgroup must be a union of conjugacy classes and must include the identity. Its order therefore has the formThe resulting distinct possibilities areOf these, only $1$ and $60$ divide $60$. Lagrange's theorem rules out every other possible order. Hence