I think that the center is ##\{R_0, R_{180} \}##. So the order of ##D_4 / Z(D_4)## is 4, which makes the order of ##Inn( D_4)## 4?fresh_42 said:You've just proven ##Inn(D_4) \cong D_4/Z(D_4)##. So what is the center of ##D_4##?
Would have been my guess, too, but this is no proof. And even if we know the order were four, which one of the two groups of order four is it? do you know a representation of ##D_4## as a semi-direct product or which are the normal subgroups of ##D_4##?Mr Davis 97 said:I think that the center is ##\{R_0, R_{180} \}##. So the order of ##D_4 / Z(D_4)## is 4, which makes the order of ##Inn( D_4)## 4?
The set of inner automorphisms of D_4, denoted by Inn(D_4), is the collection of all automorphisms of the dihedral group D_4 that are induced by conjugation by elements of D_4.
The order of Inn(D_4) can be found by counting the number of elements in D_4 that are conjugate to each other. This can be done by listing out all possible combinations of elements in D_4 and checking if they are conjugate or not.
Yes, Inn(D_4) is a subgroup of D_4. It is closed under composition, has an identity element (the identity automorphism), and every element has an inverse (the conjugate by the inverse element). Therefore, it satisfies the three conditions for being a subgroup.
The order of Inn(D_4) is equal to the index of the center of D_4 in D_4. This means that the number of elements in Inn(D_4) is equal to the number of elements in D_4 that are not in the center.
The set of inner automorphisms of D_4 is closely related to the normal subgroups of D_4. By studying the conjugacy classes of D_4, we can determine the normal subgroups and use this information to understand the structure of D_4. Additionally, inner automorphisms can be used to prove theorems and properties of D_4 by exploiting their relation to the center of D_4.