Let \(H\) be a subgroup of a finite group \(G\).
Show that \(H\) has the same number of left cosets as right cosets.
Let \(G\) be the group of symmetries of the square. Find a subgroup \(H\) and an element \(x\in G\) such that \[ xH\ne Hx. \]
Solution
1. The map\[ gH\longmapsto Hg^{-1} \]is a bijection from the set of left cosets of \(H\) to the set of rightcosets of \(H\).
For every \(g\in G\), \[ (gH)^{-1} = \{(gh)^{-1}:h\in H\} = \{h^{-1}g^{-1}:h\in H\} = Hg^{-1}, \] because inversion permutes the elements of the subgroup \(H\). Therefore the displayed assignment is simply set inversion applied to a left coset. Since inversion of subsets is an involution, this assignment is bijective.
2. The subgroup \(H\) has the same number of left cosets as rightcosets.
This follows immediately from the bijection in step 1.
3. For part 2, write the symmetry group of the square as\[ G = \langle r,s:r^4=s^2=1,\ srs=r^{-1}\rangle, \]where \(r\) is rotation through \(90^\circ\) and \(s\) is a reflection, andtake\[ H=\langle s\rangle=\{1,s\}. \]
The element \(s\) has order \(2\), so \(H=\{1,s\}\) is a subgroup of \(G\). The displayed presentation records the standard relation between a quarter-turn and a reflection of the square.
4. With \(x=r\),\[ \boxed{rH\ne Hr}. \]
We have \[ rH=\{r,rs\}, \qquad Hr=\{r,sr\}. \] From \(srs=r^{-1}\) and \(s^2=1\), \[ sr=r^{-1}s=r^3s. \] If \(rs=sr\), then multiplying on the right by \(s\) would give \[ r=r^{-1}, \] hence \(r^2=1\), contradicting that \(r\) is a quarter-turn of order \(4\). Thus \(rs\ne sr\), so the two displayed cosets are different.
5. Q.E.D.
Step 2 proves part 1, and steps 3 and 4 give the requested example for part 2.