What about groups of order 28?

Z28 is one

D14 is another

n2=1(mod2) and n2 divides 7 => n2 = 1 or 7

n7=1(mod7) and n7 divides 4 => n7 = 1

Since the 7-sylow group is unique it must be normal in G.

If n2=1 , then the 2-sylow group would be unique and normal in G too. That would make the direct product of Z4xZ7 which is isomorphic to Z28. An abelian group.

If n2=7, then there are seven 2 sylow groups. That makes one identity and 21 elements of order 4. Those 22 elements pluse the 6 elements of order 7 makes a wooping total of 28 elements in G.

Q: How do I describe this group?

I think it's three different semi direct products with three different alphas. That's the function used to multiply the left elements (usually). If the alpha is the identity in Z4 then it would just do nothing and the group would be isomorphic to Z28. If the alpha operates from one of the three other elements that makes three different isomorphic groups??

Is there some website that describes how to get the semi-direct products out of this?