One last question for the night...

**MichaelMath** · October 27th 2010, 06:49 AM

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 $C$ .

**tonio** · October 27th 2010, 10:44 AM

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

.

**MichaelMath** · October 28th 2010, 05:51 PM

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?

**tonio** · October 28th 2010, 07:48 PM

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

**MichaelMath** · October 28th 2010, 08:13 PM

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...

Thread: One last question for the night...

LinkBack

Thread Tools

Display

One last question for the night...

Similar Math Help Forum Discussions

last question of the night..thank goodness

Help me on these all night stumpers

having a slow night!

Two More Questions For The Night.

last thread for the night

Search Tags