Question

Group Theory
Could someone please explain to me why S3 is isomorphic to D6, and not to D3? Thanks :)

Answer 1

no no one can explain it i hope

Answer 2

at least in my understanding \(D_6\) has order 12 but maybe i am mistaken

Answer 3

how exactly are you representing your elements of \(S_3\) ?

Answer 4

whew i got scared for a moment

Answer 5

that may give an insight

Answer 6

you can exhibit the isomorphism directly, but it depends on what you call your elements
saying they are isomorphic amounts to saying it makes no difference, but you have to call them something

Answer 7

always a pleasure getting an insight from you, @satellite73

Answer 8

lol thanks

Answer 9

Sorry I'm confused, are you saying S3 isn't isomorphic to D6?

Answer 10

Btw this isn't homework or anything, I'm just trying to better my understanding of isomorphisms, for example, I'm still unclear as to why
Z(2)xZ(2) is isomorphic to V4?

I understand the definition, or I should rather say, I have the definition, but I am finding it difficult to see why they are isomorphic, is it to do with no of elements?

Thanks so much for any help :)

Answer 11

What's V4?

Answer 12

nvm, google :)

Answer 13

from what I've gathered, yes
but I think the key is what satellite said

Answer 14

They're isomorphic because they each have three elements of order 2?
(About \(\large \mathbb{Z}_2\times \mathbb{Z}_2 \) and \(\large V_4\) )

Answer 15

It seems odd because I've always defined the Klein-4 as the \(\large \mathbb{Z}_2 \times \mathbb{Z}_2\) itself, though I suppose it could go like this
\[\Large V_4 = \left<a,b\right>\]where\[\Large a^2 = b^2 = (ab)^2 = 1\]
Then just directly establish an isomorphism
\[\large (0,0) \rightarrow1\\\large (1,0)\rightarrow a\\\large (0,1)\rightarrow b\\\large(1,1)\rightarrow ab\]

Answer 16

Oh okay, that makes sense!
Thankyou for the help guys :)

Answer 17

"an isomorphism f gives a one-to-one correspondence between the elements of two groups, which preserves the group operation. In other words, the fact that two groups are isomorphic, means that we can simply rename the elements of the first group (calling x by f(x)) and obtain exactly the second group.

Obviously, the way we name elements is not important, so from the point of view of the group theory, isomorphic groups are identical."

Answer 18

So because both S3 and D3 have the same elements, they are isomorphic? Because you can map any element of S3 in D3..is that right?

Answer 19

injective mapping, yes

Answer 20

bijective?

Answer 21

mistyped (misspoke) bijective, yes

Answer 22

The dihedral and permutation groups make me dizzy -.-

Answer 23

you got that right. it's like one of the things I have to read and re-read

Answer 24

why can't we just go back to the old days when we only bartered LMAO

Answer 25

Haha nin ^^ so true

Answer 26

@KingGeorge

Answer 27

Do you still need any help here?

Answer 28

The only thing i'm struggling with now is what the differences are between the groups...
Like D/S/Z/V?

Answer 29

Well, \(D_{n}\) or \(D_{2n}\) depending on your notation, is the dihedral group, and basically corresponds to rotations/reflections of the \(n/2\)-gon or \(n\)-gon (depending on notation).

\(S_n\) is the symmetric group, and is usually thought of as permutations of the set \(\{1,2,...,n\}\). This group has order \(n!\) and is very nice group to look for examples in. There is in fact a theorem that states: If a finite group has order \(n\), then it is isomorphic to a subgroup of \(S_{n!}\).

\(\mathbb{Z}_n\) is simply the cyclic group of order \(n\).

Finally, \(V_4\) is the Klein group. I haven't worked with this as much so I'm not an expert on it. However, it is normal in

Answer 30

Whoops. That last sentence petered out there. It's normal in \(A_4\), which is itself normal in \(S_4\).

\(A_n\) is the alternating group, and is a subgroup of \(S_n\). It's the set of even permutations, and is generated by the 3-cycles in \(S_n\). (If you haven't learned about normality or \(A_4\) yet, ignore this post