Yes, a group of order 32 is solvable. To explain why, let's first go through the definitions and the concept of solvability.

Definition: Solvable Group

A group $G$ is called solvable if it has a subnormal series: $\{e\} = G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \dots \triangleleft G_n = G$ where each $G_i$ is a normal subgroup of $G_{i+1}$, and the quotient groups $G_{i+1}/G_i$ are abelian.

In simple terms, a group is solvable if you can build a series of normal subgroups such that the successive quotients are abelian.

Proof That a Group of Order 32 is Solvable

By the Burnside's theorem, if a group has order $p^a q^b$ (i.e., its order is the product of powers of primes), where $p$ and $q$ are primes, then the group is solvable.

In our case, the order of the group is 32, which is $2^5$. Since it's a prime power (i.e., a group whose order is a power of a single prime), we can use a key result from group theory:

• All groups of prime power order are solvable.

This is because such groups have a non-trivial center, and you can construct a normal series where the successive quotients are abelian.

Steps to Construct the Normal Series:

Let $G$ be a group of order 32. By the class equation and the properties of groups of prime power order, we know that $G$ has a non-trivial center $Z(G)$. Here's how the proof proceeds:

1. Step 1: Existence of a non-trivial center Since $G$ is a $p$-group (in this case, a 2-group), it has a non-trivial center $Z(G)$. The center $Z(G)$ is a normal subgroup of $G$ because the center of any group is normal in the group.

2. Step 2: Quotient group is of smaller order The quotient group $G/Z(G)$ has order less than 32 (specifically, order $32/|Z(G)|$). Since $Z(G)$ is non-trivial, $G/Z(G)$ has order a power of 2 (specifically, less than 32). This process can be repeated for $G/Z(G)$, since it's also a $p$-group.

3. Step 3: Continue the process We continue this process of finding normal subgroups and constructing quotients. Because the order of each successive quotient group decreases, eventually, we reach the trivial group $\{e\}$.

4. Step 4: Abelian quotients At each stage, the quotient groups formed (such as $G/Z(G)$) are abelian because we are dealing with $p$-groups and the successive centers/quotients are abelian.

5. Step 5: Finite process This is a finite process, and hence, we can construct a series of normal subgroups where the quotients are abelian.

Therefore, by following this procedure, we can form a normal series for $G$ with abelian successive quotients, showing that the group is solvable.

Conclusion:

Thus, a group of order 32 is solvable because it can be broken down into a normal series with abelian quotients, and it satisfies the conditions of Burnside's theorem for solvability.

---

Let me know if you'd like further clarification or have any questions.

Here are some related questions you might consider exploring:

1. How do we find the center of a group?
2. What is the class equation and how does it relate to group centers?
3. Can a group of prime power order ever be non-abelian?
4. What is the importance of Burnside's theorem in the theory of finite groups?
5. How do solvable groups relate to the concept of radical groups?

Tip: When proving solvability, constructing a normal series and checking the nature of the quotient groups is crucial. Always remember to use basic properties like the existence of the center in $p$-groups.