# One last question for the night...

• Oct 27th 2010, 07:49 AM
MichaelMath
One last question for the night...
Consider $S_5$, the symmetric group of degree five. Does it have a subgroup isomorphic to $C_5 \times C_5$ ? Does it have elements of order $6$? Does it have a subgroup isomorphic to $D_5$ ? What about a subgroup isomorphic to $D_6$? Let $C \subset S_5$ be a cyclic subgroup of order three. Find, up to isomorphism, the centraliser and the normaliser of $C$.
• Oct 27th 2010, 11:44 AM
tonio
Quote:

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

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

Does it have elements of order $6$?

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

Does it have a subgroup isomorphic to $D_5$ ?

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

What about a subgroup isomorphic to $D_6$?

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

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

You try this yourself.

Tonio

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• Oct 28th 2010, 06:51 PM
MichaelMath
For $S_5$ having a subgroup isomorphic to $D_6$,

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

where:

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

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

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

$e \rightarrow 1$

$r \rightarrow (12345)=\sigma$

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

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

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

Does that make sense?
• Oct 28th 2010, 08:48 PM
tonio
Quote:

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

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

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

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

$e \rightarrow 1$

$r \rightarrow \sigma$

$r^2 \rightarrow \sigma^2$

$r^3 \rightarrow \sigma^3$

$r^4 \rightarrow \sigma^4$

Does that make sense?

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

Tonio
• Oct 28th 2010, 09:13 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...