Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 3 / 5D (Groups)

The order of an element $x$ is the least positive integer $d$ such that $x^d=e$, and the order of a finite group is its number of elements.

Lagrange theorem states that if $H\le G$ and $G$ is finite, thenIndeed, the left cosets of $H$ partition $G$. Multiplication by a coset representative is a bijection from $H$ to each coset, so every coset has $|H|$ elements. Summing over the $[G:H]$ cosets proves the formula.

The cyclic subgroup $⟨x⟩$ has exactly $d$ elements when $x$ has order $d$. Applying Lagrange to $⟨x⟩\le G$ gives

No. The alternating group $A_4$ has order twelve, and six is a proper divisor of twelve, but $A_4$ has no element of order six. Its nonidentity elements are eight 3-cycles of order three and three double transpositions of order two.

Let $a$ and $b$ generate $C_m$ and $C_n$. The order of $(a,b)$ in the direct product of groups isWhen $\gcd (m,n)=1$, this is $mn=|C_m\times C_n|$. Hence $(a,b)$ generates the whole product and

Every element satisfies $x^2=e$, including the identity. For any $x,y\in G$,Since $x^{-1}=x$ and $y^{-1}=y$, this implies $xy=yx$, so $G$ is abelian. It is therefore a vector space over the two-element field, with group operation as vector addition and scalar multiplication by zero or one. A basis of this finite vector space givesfor some $r\ge 0$.

Write $d=qr+s$ with $0\le s<r$. Since $x^d=e\in H$ and $x^r\in H$,The minimality of the positive integer $r$ forces $s=0$. Therefore

Among the $n+1$ right cosetstwo must coincide because $[G:H]=n$. If $H{x}^i=H{x}^j$ with $0\le i<j\le n$, right multiplication by $x^{-i}$ gives $H=H{x}^{j-i}$, so

The least such exponent need not divide $n$. For example, take $G=S_3$, $H=⟨(12)⟩$, and $x=(13)$. Then $[G:H]=3$, while $x\notin H$ and $x^2=e\in H$, so the least exponent is two, which does not divide three.