# One last question for the night...

• Oct 27th 2010, 06:49 AM
MichaelMath
One last question for the night...
Consider $\displaystyle S_5$, the symmetric group of degree five. Does it have a subgroup isomorphic to $\displaystyle C_5 \times C_5$ ? Does it have elements of order $\displaystyle 6$? Does it have a subgroup isomorphic to $\displaystyle D_5$ ? What about a subgroup isomorphic to $\displaystyle D_6$? Let $\displaystyle C \subset S_5$ be a cyclic subgroup of order three. Find, up to isomorphism, the centraliser and the normaliser of $\displaystyle C$.
• Oct 27th 2010, 10:44 AM
tonio
Quote:

Originally Posted by MichaelMath
Consider $\displaystyle S_5$, the symmetric group of degree five. Does it have a subgroup isomorphic to $\displaystyle C_5 \times C_5$ ?

Of course not: $\displaystyle |C_5\times C_5|=25\nmid 120=5!$ , so by Lagrange it can't be

Does it have elements of order $\displaystyle 6$?

Lots of: any element of the form $\displaystyle (ab)(cde)$ , with different a,b,c,d,e has order 6

Does it have a subgroup isomorphic to $\displaystyle D_5$ ?

Check the subgroup generated by $\displaystyle (14)(23),\,(12345)$

What about a subgroup isomorphic to $\displaystyle D_6$?

Check the subgroup $\displaystyle <(34),\,(12)(345)>$

Let $\displaystyle C \subset S_5$ be a cyclic subgroup of order three. Find, up to isomorphism, the centraliser and the normaliser of $\displaystyle C$.

You try this yourself.

Tonio

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• Oct 28th 2010, 05:51 PM
MichaelMath
For $\displaystyle S_5$ having a subgroup isomorphic to $\displaystyle D_6$,

Can I use the group of rotations in $\displaystyle D_6$: $\displaystyle r^n \, ( n =1,2,3,4,5)$, and $\displaystyle \sigma = (1,2,3,4,5)$ in $\displaystyle S_5$

where:

$\displaystyle X = \{ e,r,r^2,r^3,r^4 \}$

$\displaystyle Y = (1,\sigma, \sigma^2,\sigma^3,\sigma^4)$ in $\displaystyle S_5$

Then the isomorphism $\displaystyle \phi: X \rightarrow Y$ is:

$\displaystyle e \rightarrow 1$

$\displaystyle r \rightarrow (12345)=\sigma$

$\displaystyle r^2 \rightarrow (13524)=\sigma^2$

$\displaystyle r^3 \rightarrow (14253)=\sigma^3$

$\displaystyle r^4 \rightarrow (15432)=\sigma^4$

Does that make sense?
• Oct 28th 2010, 07:48 PM
tonio
Quote:

Originally Posted by MichaelMath
For $\displaystyle S_5$ having a subgroup isomorphic to $\displaystyle D_6$, is the subgroup of $\displaystyle D_6$, the group of rotations?

So if the subgroup of rotations $\displaystyle X = \{ e,r,r^2,r^3,r^4 \}$

and $\displaystyle \sigma = (1,2,3,4,5)$ in $\displaystyle S_5$

Then the isomorphism $\displaystyle \phi: X \rightarrow S_5$ is:

$\displaystyle e \rightarrow 1$

$\displaystyle r \rightarrow \sigma$

$\displaystyle r^2 \rightarrow \sigma^2$

$\displaystyle r^3 \rightarrow \sigma^3$

$\displaystyle r^4 \rightarrow \sigma^4$

Does that make sense?

Caution: Most people denote by $\displaystyle D_n$ the dihedral group of order 2n , not the one of order n.

Tonio
• Oct 28th 2010, 08:13 PM
MichaelMath
I made a few changes above...

arghgh...

So out of (12)(35), (13,45), (15)(24), (25)(34), and (14,23), the last one fixes 5... does this mean anything...