Answer: G = (19) × (26) × (16)
Step-by-step explanation:
The isomorphism classes of Abelian groups of order 12 are Z₄ ⊕ Z₃ and Z₂ ⊕ Z₂ ⊕ Z₃
SO Let us calculate the orders of some of the elements of G
We have
4² = 16,
4³ = 64
= 19,
and
4^4 = 19.4
= 76
= 31.
furthermore,
4^5 = 31.4
= 124
= 34
and
4^6 = 34.4
= 136
= 1
Hence, 4 and 34 each have order 6, 16 and 31 each have order 3, and 19 has order 2.
Next, we calculate
11² = 121
= 31
and
11³ = 11.3
= 341
= 26
this is the calculation needed.
26² = 11^6
= 31³
= 1
since we already showed that 31 has order 3. This means that 26 has order 2
Since G has two distinct elements of order 2, it cannot be isomorphic to . We conclude
that G = Z₂ ⊕ Z₂ ⊕ Z₃
Finally, we will express as an internal direct product.
The previous calculations show that
(19) = { 1, 19 }
and (26) = { 1, 26 }
are cyclic subgroups of G of order 2 with trivial intersection. We have
(19) × (26) = { 1, 19, 26, 44 }
since
(16) = { 1, 19, 26, 44 }
has trivial intersection with (19) × (26), conclude that
G = (19) × (26) × (19)
