Question: Suppose that G = H x K and that N is a normal subgroup of H. Prove that N is normal in G.
This doesn't appear to be too difficult, but for some reason I'm stuck. Since G = H x K, we can write any g in G as hk. So Ng = N(hk) = hNk, and then after that I'm not sure what to do. I'm not even sure if that is the right general path. Anyway, any help is appreciated.
This doesn't appear to be too difficult, but for some reason I'm stuck. Since G = H x K, we can write any g in G as hk. So Ng = N(hk) = hNk, and then after that I'm not sure what to do. I'm not even sure if that is the right general path. Anyway, any help is appreciated.
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Let N be a normal subgroup of H and let G = H x K. That is, G is the direct product of the groups H and K, consisting of elements of the form (h,k) with multiplication:
(x,y)(x',y') = (xx',yy').
The identity element is (e_H,e_K), and the inverse of any element (x,y) is (x^-1,y^-1).
You seem to have mistaken the direct product of two groups with the quotientgroup!
I will show you a general proof. So N is a normal subgroup of H. This means that it is a subgroup which is closed under conjugation of its elements:
hnh^-1 is in N for all h in H and n in N.
(equivalently, a subgroup N is normal iff Nh=hN for all h in H).
Now we have to show that N is also a normal subgroup of G. However, this should strike you as odd. In order for N to be normal in G, it would have to consist of pairs of elements (in particular, pairs of elements of the form (h,k) where h is in H and k in K), but it only consists of single elements in H.
However, we can manufacture a normal subgroup in G=HxK isomorphic to N. In that sense we can say that N is normal in G. The group in G to consider is Nx{e_H}. This subgroup is isomorphic to N, and it is almost trivial to show that it is normal in G:
(h,k) (n,e)(h^-1,k^-1) = (hnh^-1,kek^-1) = (hnh^-1,e) = (n',e).
Here I've used the multiplication in the direct product and that the inverse of a given element (h,k) is (h^-1,k^-1) for conjugation. Also I've used the assumption that N is normal, which tells us that hnh^-1 = n' for some n' in N. Now we see that Nx{e} is a normal subgroup in G.
This normal subgroup is isomorphic to N, because there is an isomorphism f(n) = (n,e_H) from N to Nx{e}.
(x,y)(x',y') = (xx',yy').
The identity element is (e_H,e_K), and the inverse of any element (x,y) is (x^-1,y^-1).
You seem to have mistaken the direct product of two groups with the quotientgroup!
I will show you a general proof. So N is a normal subgroup of H. This means that it is a subgroup which is closed under conjugation of its elements:
hnh^-1 is in N for all h in H and n in N.
(equivalently, a subgroup N is normal iff Nh=hN for all h in H).
Now we have to show that N is also a normal subgroup of G. However, this should strike you as odd. In order for N to be normal in G, it would have to consist of pairs of elements (in particular, pairs of elements of the form (h,k) where h is in H and k in K), but it only consists of single elements in H.
However, we can manufacture a normal subgroup in G=HxK isomorphic to N. In that sense we can say that N is normal in G. The group in G to consider is Nx{e_H}. This subgroup is isomorphic to N, and it is almost trivial to show that it is normal in G:
(h,k) (n,e)(h^-1,k^-1) = (hnh^-1,kek^-1) = (hnh^-1,e) = (n',e).
Here I've used the multiplication in the direct product and that the inverse of a given element (h,k) is (h^-1,k^-1) for conjugation. Also I've used the assumption that N is normal, which tells us that hnh^-1 = n' for some n' in N. Now we see that Nx{e} is a normal subgroup in G.
This normal subgroup is isomorphic to N, because there is an isomorphism f(n) = (n,e_H) from N to Nx{e}.