# normal subgroup question

#### nhk

Let H and K be normal Subgroups of a group G s.t H intersect K = {e}. SHow that G is isomorphict to a subgroup of G/H + G/K.
Proof/
Lets define are mapping f:G to G/H+G/K by f(g)=(gH,GK). Need to show that this is a homomorphism. Obviously f is a well defined function. Operation Preserving- Let x,y be elements of G. Then f(xy)= (xyH,xyK)= (xHyH,xKyK)= (xH,XK),(yH,yK)=f(x)f(y) ( I need help justifying my step on this)
Thus f is a homomorphism. Since kerf ={g element of G| (gH,gK)=(H,K)}. thus f is injective (one to one).
So By 1st isomorphism theroem, G/Ker f is isomorphic to f(G). Since f(G) is a subgroup of G/H+G/K (since f is one to one), then G is isomorphic to a subgroup of G/H + G/K.
Does this seem right?

#### tonio

Let H and K be normal Subgroups of a group G s.t H intersect K = {e}. SHow that G is isomorphict to a subgroup of G/H + G/K.
Proof/
Lets define are mapping f:G to G/H+G/K by f(g)=(gH,GK). Need to show that this is a homomorphism. Obviously f is a well defined function. Operation Preserving- Let x,y be elements of G. Then f(xy)= (xyH,xyK)= (xHyH,xKyK)= (xH,XK),(yH,yK)=f(x)f(y) ( I need help justifying my step on this)
Thus f is a homomorphism. Since kerf ={g element of G| (gH,gK)=(H,K)}. thus f is injective (one to one).
So By 1st isomorphism theroem, G/Ker f is isomorphic to f(G). Since f(G) is a subgroup of G/H+G/K (since f is one to one), then G is isomorphic to a subgroup of G/H + G/K.
Does this seem right?

1) $$\displaystyle (xHyH,\,xKyK)=(xH,\,xK)(yH,\,yK)$$ because this is the definition of product in a direct product

2) You wrote "...thus f is injective". How "thus"?? For this you must prove that $$\displaystyle \ker f=\{1\}$$ , which of course follows at once from the given data, but you haven't yet pointed this out.

The rest seems to be fine.

Tonio

• nhk