Intersection of Normal groups

• Oct 9th 2007, 07:21 PM
Intersection of Normal groups
Let H and K be normal subgroups of G, prove that H intersect K is also normal in G.

Proof:

Since H and K are normal in G, we have

$aH=Ha \forall a \in G$

$bK=Kb \forall b \in G$

Now, let $c \in H \cap K$, then c must retain the property of H and K since c is in both.

Thus, $a(c)=(c)a \forall a \in G$, implies $a(H \cap K) = (H \cap K)a \forall a \in G$

Therefore H intersect K is normal in G.

Is this right?
• Oct 10th 2007, 10:54 AM
Opalg
Quote:

Let H and K be normal subgroups of G, prove that H intersect K is also normal in G.

Proof:

Since H and K are normal in G, we have

$aH=Ha \forall a \in G$

$bK=Kb \forall b \in G$

Now, let $c \in H \cap K$, then c must retain the property of H and K since c is in both.

Thus, $a(c)=(c)a \forall a \in G$, implies $a(H \cap K) = (H \cap K)a \forall a \in G$

Therefore H intersect K is normal in G.

Is this right?

No it's not right. The problem is that you are confusing the equation $a(H \cap K) = (H \cap K)a$ with the equation $ac=ca$ where c is an element of $H\cap K$.

Looking at H rather than $H\cap K$ for a moment, the equation $aH=Ha$ means that the sets aH and Ha have the same elements. This does not mean that ah=ha for each element h of H. Instead, it means that, given h in H, ah=h'a for some element h' of H (which may be different from h).

Now coming back to the subgroup $H\cap K$, it is not true that if $c \in H \cap K$ then $a(c)=(c)a\; (\forall a \in G)$. What you can say is that $a(H\cap K)\subseteq aH=Ha$ and $a(H\cap K)\subseteq aK=Ka$. You then need to show that $a(H\cap K)\subseteq (Ha)\cap(Ka)=(H\cap K)a$. A similar argument will then show the reverse inequality $(H\cap K)a\subseteq a(H\cap K)$.
• Oct 10th 2007, 05:26 PM
ThePerfectHacker
Again, use my version of the equivalent condition for being normal.
$ghg^{-1}\in H \mbox{ and }khk^{-1}\in K$ for all $h\in H, k\in K, g\in G$.

Now let $x\in (H\cap K)$ it means $x\in H\mbox{ and }x\in K$. So $gxg^{-1} \in H \mbox{ and }gxg^{-1} \in K$ by normality. Thus, $gxg^{-1} \in (H\cap K)$. Thus, $(H\cap K)\triangleleft G$. Q.E.D.

My point is you should learn the way I do it. My approach certainly does not look as elegant as saying the left-right cosets. But it is much more efficient because I am using single elements only while you are working with an entire set.