Is N3 a Subgroup of Dihedral Group Dih(12)?

[facepalmer]

Homework Statement

Taking the Dih(12) = {α,β :α⁶ = 1, β² = 1, βα = α^-1β}
and a function N_r = {g^r: g element of Dih(12)}

Homework Equations

Taking the above I have to find the elements of N₃. And then prove that N₃ is not a subgroup of Dih(12).

The Attempt at a Solution

For N₃ I have determined the following are elements. 1, α³, β, βα³.
Now to attempt to prove whether these elements form a subgroup of Dih(12) I created the multiplication table. The result was the table was closed and so I would assume that the aforementioned subset is a subgroup.
I assume I have done one of two things wrong; either
a) My understanding of generating n₃ is misguided, though if I take an element of Dih(12), say αβ, and then using the function defined I get (αβ)³ = α³ββ = α³.

or b) my multiplication table is wrong.

Any help on what my mistake is is appreciated.

 

[micromass]

(αβ)³ = α³ββ = α³.

I don't get how you did this calculation. Could you show more steps??

 

[conquest]

I think you did (αβ)^3 = α^3β^3 but actually (αβ)^3=αβαβαβ

so say (αβ)^3= g then αβαβαβ=g so βαβαβ=α^-1g

thus βαβαβ=βαβg so g=αβ

 

[facepalmer]

in reply to micromass...
(σβ)³ I did: αβαβαβ = αααβββ = α³β³ = α³β (as ββ = 1)
Now, if I follow what conquest has said I should have (and recalling βα = α-1β)
(σβ)³ = αβαβαβ = αα^-1α^-1β = 1α^-1β = βα.
and for some of the other elements I get
(α²β)³ = α²βα²βα²β = α²α^-2ββα²β = α²β
(α³β)³ = α³βα³βα³β = α³α^-3ββα³β = α³β

therefore If I apply the same understanding as above to every element of Dih(12) I get the following results for N₃
...
So the elements of N₃ are
1, α³, β, βα, βα², βα³, βα⁴, βα⁵

and if I do out the multiplication table all elements of N₃ are contained in the table, along with the following elements that are not elements of the set N₃
α, α², α⁴, α⁵
hence N3 is not a subgroup of Dih(12).

Am I going on the right tracks with this?

 

[facepalmer]

I should also add that for
(α)³ = ααα = α³
(α²)³ = α²α²α² = α⁶ = 1
(α³)³ = α³α³α³ = α⁹ = α⁶α³ = α³
(α⁴)³ = α⁴α⁴α⁴ = α¹² = α⁶α⁶ = 1
(α⁵)³ = α⁵α⁵α⁵ = α¹⁵ = α¹²α³ = α³

 

[conquest]

exactly for instance:

β³=β
(αβ)³=αβ

and βαβ=α^-1= α⁵

which is not in N₃

 

[facepalmer]

Excellent, many thanks!