Important: The group package has been deprecated. Use the superseding command GroupTheory[IsNormal] instead.

With two arguments, this function determines whether sg is a normal subgroup of the permutation group generated by the union of sg and pg.  If it is, then true is returned; otherwise false is returned.

With one argument, this function tests whether subgrl is a normal subgroup.  The function returns true or false.

The command with(group,isnormal) allows the use of the abbreviated form of this command.

$\mathrm{with}\left(\mathrm{group}\right)\:$

$\mathrm{pg}≔\mathrm{permgroup}\left(8\,\left\{\left[\left[1\,2\right]\right]\,\left[\left[1\,2\,3\,4\,5\,6\,7\,8\right]\right]\right\}\right)\:$

$\mathrm{sg1}≔\mathrm{permgroup}\left(8\,\left\{\left[\left[1\,2\right]\right]\,\left[\left[5\,6\right]\right]\,\left[\left[1\,2\,3\,4\right]\right]\,\left[\left[5\,6\,7\,8\right]\right]\right\}\right)\:$

$\mathrm{isnormal}\left(\mathrm{pg}\,\mathrm{sg1}\right)$

$\mathrm{false}$

$\mathrm{sg2}≔\mathrm{permgroup}\left(8\,\left\{\left[\left[1\,2\,3\right]\right]\,\left[\left[2\,3\,4\,5\,6\,7\,8\right]\right]\right\}\right)\:$

$\mathrm{isnormal}\left(\mathrm{pg}\,\mathrm{sg2}\right)$

$\mathrm{true}$

$g≔\mathrm{grelgroup}\left(\left\{a\,b\right\}\,\left\{\left[b\,b\,b\right]\,\left[a\,b\,\frac{1}{a}\,\frac{1}{b}\right]\,\left[a\,a\,a\,a\,a\right]\right\}\right)\:$

$\mathrm{isnormal}\left(\mathrm{subgrel}\left(\left\{x=\left[a\right]\right\}\,g\right)\right)$

$\mathrm{true}$