Normal subgroup

**tttcomrader** · October 8th 2007, 03:03 PM

Let G be a group, and let $H \leq G$

(That means H is the normal subgroup of G, right?)

a)Show that $aHa^{-1}$ is also a subgroup of G.

My proof:

Suppose that x,y is in G, so that $xHx^{-1}, yHy^{-1} \in aHa^{-1}$

Since H is normal in G, pick $Hx^{-1}=x_{1}H, yH=Hy_{1},x_{1},y_{1} \in G$

$(xHx^{-1})(yHy^{-1}) = x(x_{1}H)(Hy_{1}y^{-1}) = xx_{1}Hy_{1}y^{-1}$

Now, $(xHx^{-1})^{-1}=x^{-1}Hx \in aHa^{-1}$

Is this right?

b) If H is finite of order k, then $aHa^{-1}$ is also finite of order k.

**Plato** · October 8th 2007, 03:50 PM

Do you know this theorem?
A set H in a group G is a subgroup iff $a \in H\;\& \;b \in H \Rightarrow \quad ab^{ - 1} \in H$ ?

Suppose that $x \in aHa^{ - 1} \;\& \;y \in aHa^{ - 1} \Rightarrow \left( {\exists c \in H} \right)\left( {\exists d \in H} \right)\left[ {x = aca^{ - 1} \wedge y = ada^{ - 1} } \right]$ .

What can be said about $xy^{ - 1}$ ?

**ThePerfectHacker** · October 8th 2007, 05:11 PM

Originally Posted by tttcomrader

b) If H is finite of order k, then $aHa^{-1}$ is also finite of order k.

$\{ a_1, ... , a_n \}$
Let the above be the elements of $H$ .
Then,
$\{aa_1a^{-1},...,aa_na^{-1} \}$
Are all distint.
Because otherwise if,
$aa_ia^{-1} = aa_ja^{-1}$
This implies,
$a_i = a_j$ .
Which is an impossibility for distinct elements.
-----
Here is a useful theorem to know.

Theorem: Let $H$ be a subgroup of $G$ the following are equivalent.
1)For all $g\in G$ and $h\in H$ implies $ghg^{-1} \in H$ .
2)For all $g\in G$ we have $gHg^{-1}$ .
3)The left and right cosets coincide, i.e. $aH = Ha$ .

All these 3 are equivalent formulations for "normal subgroup". My favorite is #1.

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