To determine if any group of order 12 is simple, let's analyze the structure of groups with order 12 using the Sylow theorems.

Step 1: Factorization of the Order

The order of the group $|G| = 12$ can be factored as: $12 = 2^2 \times 3$ This means that any group of order 12 must have Sylow $2$-subgroups of order $2^2 = 4$ and Sylow $3$-subgroups of order $3$.

Step 2: Applying Sylow's Theorems

1. Number of Sylow 3-subgroups:

   • Let $n_3$ be the number of Sylow 3-subgroups.
   • By Sylow's theorems, $n_3$ must divide $|G|$ and $n_3 \equiv 1 \ (\text{mod} \ 3)$.
   • Since $n_3$ divides 4 (the quotient $12/3 = 4$), possible values for $n_3$ are 1 or 4.

2. Number of Sylow 2-subgroups:

   • Let $n_2$ be the number of Sylow 2-subgroups.
   • By Sylow's theorems, $n_2$ must divide $|G|$ and $n_2 \equiv 1 \ (\text{mod} \ 2)$.
   • Since $n_2$ divides 3 (the quotient $12/4 = 3$), possible values for $n_2$ are 1 or 3.

Step 3: Analyzing Simplicity

• Case 1: $n_3 = 1$:

  • If there is exactly one Sylow 3-subgroup, it must be normal in $G$. This means $G$ has a non-trivial normal subgroup and hence is not simple.

• Case 2: $n_3 = 4$:

  • If there are four Sylow 3-subgroups, then each Sylow 3-subgroup has an order of 3 and the intersection of any two of them is just the identity element (since any two different Sylow $p$-subgroups intersect trivially).
  • The group has a total of $4 \times 2 = 8$ elements in these subgroups, leaving 4 elements for the Sylow 2-subgroups.

• Case 3: $n_2 = 1$:

  • If there is exactly one Sylow 2-subgroup, it must be normal in $G$. Therefore, $G$ would again have a non-trivial normal subgroup and is not simple.

• Case 4: $n_2 = 3$:

  • If there are three Sylow 2-subgroups, each of order 4, they together account for $3 \times 3 = 9$ elements. With the Sylow 3-subgroups contributing 3 elements, we again find that there is a non-trivial normal subgroup.

Conclusion

In every case, a non-trivial normal subgroup exists, implying that no group of order 12 can be simple.

So, no simple group of order 12 exists.

Would you like more details, or do you have any questions about this process?

---

Related Questions:

1. What are Sylow theorems?
2. Can you provide examples of simple groups?
3. How are Sylow subgroups identified in a group?
4. What is the significance of normal subgroups in group theory?
5. How does factorization help in group theory?
6. What are some examples of non-simple groups of order 12?
7. How do you prove that a group is simple?
8. What are cyclic groups and their properties?

Tip: When determining simplicity, always check for normal subgroups—existence implies the group isn't simple!