Yes, you are absolutely correct! The alternating group $A_5$ of order 60 is indeed the unique finite simple group of that order.

Here’s a bit of context:

• $A_5$ is the group of even permutations of five elements, which is a subgroup of the symmetric group $S_5$.
• The order of $A_5$ is given by $60$, because the symmetric group $S_5$ has order $5! = 120$, and $A_5$ is half of that, consisting of the even permutations.

Why is $A_5$ simple?

A finite simple group is one that does not have any non-trivial normal subgroups (subgroups that are invariant under conjugation). $A_5$ is simple because:

• It has no non-trivial normal subgroups.
• It is also the smallest non-abelian simple group.

This makes it a fundamental object in group theory, and it's often studied in the context of both abstract algebra and the classification of finite simple groups.

Would you like to explore further details about $A_5$, its properties, or related topics?

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Related questions for further exploration:

1. How does the structure of $A_5$ compare with that of other simple groups?
2. Why is $A_5$ considered the smallest non-abelian simple group?
3. What are some applications of $A_5$ in geometry or other areas of mathematics?
4. Can you explain the concept of even and odd permutations and their role in $A_5$?
5. What are the Sylow subgroups of $A_5$?

Tip: When studying simple groups, it's important to also understand the role of normal subgroups and conjugacy classes. These concepts often provide insight into the group's structure and its behavior under various operations.