# Normal Subgroup

• Oct 22nd 2009, 04:53 PM
ux0
Normal Subgroup
Define W= <(1 2)(3 4)>, the cyclic subgroup of $S_4$ generated by (1 2)(3 4). Show that W is a normal subgroup of V, where V ={(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}, but that W is not a normal subgroup of $S_4$. Conclude that normality is not transitive: K is normal to H, and H is normal to G does not imply K is normal to G.
• Oct 22nd 2009, 05:22 PM
Defunkt
Quote:

Originally Posted by ux0
Define W= <(1 2)(3 4)>, the cyclic subgroup of $S_4$ generated by (1 2)(3 4). Show that W is a normal subgroup of V, where V ={(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}, but that W is not a normal subgroup of $S_4$. Conclude that normality is not transitive: K is normal to H, and H is normal to G does not imply K is normal to G.

Hi!

V is also otherwise known as the Klein-four-group(!)

We know that W is normal if $\forall g \in G, w \in W , gwg^{-1} \in W$

However since the elements of W are only (2)(2) circles and so are the elements of V, and by the fact that $\forall \sigma,\rho \in S_n$ such that $\rho = (\rho(1), \rho(2),...\rho(n))$, we know that: $\sigma \rho \sigma^{-1} = (\sigma(\rho(1)), \sigma(\rho(2)),...,\sigma(\rho(n)))$

Specifically, this gives us that the circle-form of $\sigma \rho \sigma^{-1}$ is determined solely by that of $\rho$. This gives us that W is normal since the result is always a 2-2 circle and all of them are contained in the klein group.

To do the second part simply choose some other element $\xi \in S_n$ and show that $\xi \rho \xi^{-1} \notin W$
• Oct 22nd 2009, 07:19 PM
ux0
Quote:

Originally Posted by Defunkt
Hi!

However since the elements of W are only (2)(2) circles and so are the elements of V,

Just to clarify some things, this means that the elements of W are of order 2 and cyclic?.. I'm not really sure what you mean by circles we haven't discussed this in these terms in class.
• Oct 22nd 2009, 09:05 PM
tonio
Quote:

Originally Posted by ux0
Define W= <(1 2)(3 4)>, the cyclic subgroup of $S_4$ generated by (1 2)(3 4). Show that W is a normal subgroup of V, where V ={(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}, but that W is not a normal subgroup of $S_4$. Conclude that normality is not transitive: K is normal to H, and H is normal to G does not imply K is normal to G.

What've you done so far? Where're you stuck? V is what kind of group? Thus any of its subgroup is...
What about conjugating $(12)(34)\,\,by\,\,(13)?$

Tonio
• Oct 23rd 2009, 05:52 AM
ux0
Quote:

Originally Posted by tonio
What've you done so far?

Not much we are using "First course in Abstract Algebra, by Rotman, we are are Homomoprhism, all we have covered is some set theory, number theory, and a few groups.

Quote:

Originally Posted by tonio
Where're you stuck?

Im stuck on this proof because i don't understand what he means by (2)(2) circles, or later in the proof when he says 2-2 circles.... Is he saying that V has index 2 in a group G? where $S_4 = G$.

Quote:

Originally Posted by tonio
V is what kind of group? Thus any of its subgroup is...

I know V is a normal subgroup of $S_4$ ... DOes that imply that if H is a subgroup of index 2 in a group G, then H is a normal subgroup of G?

Quote:

Originally Posted by tonio
What about conjugating $(12)(34)\,\,by\,\,(13)?$

$(13) (12)(34) (31) = (3 2) (3 4)$ ??

i'm not really sure on how to do that?
• Oct 23rd 2009, 06:57 AM
Swlabr
I'm going to start from the beginning, because I can't understand the above post...this may not actually be helpful, but I can but try...

Quote:

Originally Posted by ux0
Define W= <(1 2)(3 4)>, the cyclic subgroup of $S_4$ generated by (1 2)(3 4). Show that W is a normal subgroup of V, where V ={(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}, but that W is not a normal subgroup of $S_4$. Conclude that normality is not transitive: K is normal to H, and H is normal to G does not imply K is normal to G.

Firstly, note that $V$ has order 4 and so is abelian (all groups of order 4 are abelian - there are only two of them). Thus, $W \lhd V$, because subgroups of abelian groups are always normal.

However, $W \not\lhd S_4$ because you can find an element $\sigma \in S_4$ such that $\sigma^{-1}(12)(34)\sigma \notin W$ and so $W$ cannot be normal in $S_4$.

Such a $\sigma$ is $\sigma=(13)$. $(13)(12)(34)(13)=(14)(23) \notin W$. Do you know how to perform such calculations? You go from left to right, following where each number sends you. For instance, start with 1. $1 \rightarrow 3 \rightarrow 4$. Then plugging in 4 we get $4 \rightarrow 3 \rightarrow 1$, and so we have found the first part of the cycle, (14). Then do the same with 2, and you find it goes to 3, and then we know 3 must go to 2.
• Oct 23rd 2009, 07:37 AM
Defunkt
Just to clarify, a 2-2 circle is a permutation of the form $(a b)(c d)$. What I said was that the result of $\sigma \rho \sigma^{-1}$ where $\rho$ is a 2-2 circle will always be a 2-2 circle. However, all 2-2 circles are contained in V, so this gives that W is normal. Swlabr's way is alot easier to understand, though.
• Oct 23rd 2009, 08:20 AM
ux0
Quote:

Originally Posted by Swlabr
.
$\sigma=(13)$. $(13)(12)(34)(13)=(14)(23) \notin W$.

Shouldn't $\sigma^{-1}=(31)$ therefore making the $(31)(12)(34)(13)=(12)(14)=(2 4) \notin W$???
• Oct 23rd 2009, 08:38 AM
Swlabr
Quote:

Originally Posted by ux0
Shouldn't $\sigma^{-1}=(31)$ therefore making the $(31)(12)(34)(13)=(12)(14)=(2 4) \notin W$???

Yes and no to your first question: $(13)=(31)$ (put them both into a proper permutation to see why - they both move 1 to 3 and 3 to 1 and leave everything else unchanged).

No to your second part - your calculation is incorrect. For starters, you should always end up with disjoint cycles after performing the multiplication the first time. You round off the brackets if and only if you get back to the beginning (as in, you go $a_1 \rightarrow a_2 \rightarrow \ldots \rightarrow a_i \rightarrow a_1$ and so your cycle will be $(a_1 a_2 \ldots a_i)$).