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Thread: Intersection of Normal groups

  1. #1
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    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

    $\displaystyle aH=Ha \forall a \in G$

    $\displaystyle bK=Kb \forall b \in G$

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

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

    Therefore H intersect K is normal in G.

    Is this right?
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    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

    $\displaystyle aH=Ha \forall a \in G$

    $\displaystyle bK=Kb \forall b \in G$

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

    Thus, $\displaystyle a(c)=(c)a \forall a \in G$, implies $\displaystyle 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 $\displaystyle a(H \cap K) = (H \cap K)a$ with the equation $\displaystyle ac=ca$ where c is an element of $\displaystyle H\cap K$.

    Looking at H rather than $\displaystyle H\cap K$ for a moment, the equation $\displaystyle aH=Ha$ 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 $\displaystyle H\cap K$, it is not true that if $\displaystyle c \in H \cap K$ then $\displaystyle a(c)=(c)a\; (\forall a \in G)$. What you can say is that $\displaystyle a(H\cap K)\subseteq aH=Ha$ and $\displaystyle a(H\cap K)\subseteq aK=Ka$. You then need to show that $\displaystyle a(H\cap K)\subseteq (Ha)\cap(Ka)=(H\cap K)a$. A similar argument will then show the reverse inequality $\displaystyle (H\cap K)a\subseteq a(H\cap K)$.
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  3. #3
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    Again, use my version of the equivalent condition for being normal.
    $\displaystyle ghg^{-1}\in H \mbox{ and }khk^{-1}\in K$ for all $\displaystyle h\in H, k\in K, g\in G$.

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