Let be a finite group and suppose that , are two subgroups of such that and . Show that

**Solution**

Recall that and thus . Hence and so

(1)

where and .

Now, since and we have and that is and . So if we let and then and thus

due to .

The exercise along its solution have been migrated from here .

Note:The upper bound cannot be improved; there exists a group and subgroups , of such thatAn example is the Klein-four group and the subgroups and . Then and