Yes, every group of order 32 is solvable. In fact, every group of order $p^n$, where $p$ is a prime and $n$ is a positive integer, is solvable. This is a consequence of a more general result: all $p$-groups are solvable.

Proof Using Normal Series:

To prove that a group of order 32 is solvable, we will show that it has a normal series with abelian factor groups. Here’s a step-by-step proof:

Step 1: Existence of a normal subgroup in a $p$-group

By the Sylow theorems, we know that any group of order $p^n$ (a $p$-group) has a non-trivial center, $Z(G)$, and this center is non-trivial because in a $p$-group, the center cannot be trivial. In particular, the center $Z(G)$ contains at least one non-identity element.

Since $Z(G)$ is non-trivial, it must contain a normal subgroup of $G$. Thus, $G$ contains a normal subgroup of order $p$ (i.e., the smallest prime divisor of the group's order).

Step 2: Constructing a normal series

Let $G$ be a group of order 32. We will show that we can construct a normal series for $G$ with abelian factor groups. The order of $G$ is $32 = 2^5$, so $G$ is a 2-group.

1. First Step: By the Sylow theorem, $Z(G) \neq \{e\}$. Let $H_1 = Z(G)$, which is a normal subgroup of $G$. Since $H_1$ is abelian (because it's part of the center), this gives us a normal subgroup. Moreover, $H_1$ has order greater than 1, say $|H_1| = 2^k$, where $k \geq 1$.

2. Factor group: Consider the factor group $G / H_1$. The order of this factor group is $2^{5-k}$. If $G / H_1$ is abelian, we are done because we can form a normal series $\{e\} \subset H_1 \subset G$.

   However, if $G / H_1$ is not abelian, we apply the process recursively.

3. Next Step: $G / H_1$ is still a $2$-group, so by the same argument (Sylow theorems), $Z(G/H_1)$ is non-trivial. Therefore, we can find a non-trivial normal subgroup in $G / H_1$, say $H_2/H_1$, which is abelian. Thus, $H_2$ is a normal subgroup of $G$, and we have a new normal series: $\{e\} \subset H_1 \subset H_2 \subset G$ where the factor groups are abelian.

4. Continue recursively: Repeat this process until we reach the trivial group. This process will terminate in finitely many steps because the order of the group is finite.

Step 3: Conclusion

We have constructed a normal series: $\{e\} \subset H_1 \subset H_2 \subset \dots \subset G$ where each factor group $H_{i+1}/H_i$ is abelian. Therefore, $G$ is solvable.

Summary

• Every group of order 32 is a $2$-group.
• $p$-groups (including 2-groups) have non-trivial centers.
• By constructing a normal series where the factor groups are abelian, we conclude that every group of order 32 is solvable.

Related Questions:

1. What is the definition of a solvable group?
2. How does the structure of $p$-groups ensure solvability?
3. Can we provide an example of a group of order 32 and demonstrate its solvability step by step?
4. How do the Sylow theorems guarantee the existence of normal subgroups in $p$-groups?
5. What is the role of the center of a group in proving solvability?

Tip:

Whenever you deal with groups of prime power order, remember that these groups always have a non-trivial center, which plays a crucial role in proving solvability.