Is group of order 32 solvable if yes then prove with the help of normal series .

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.

Is a group of order 32 solvable? If yes, prove it with the help of a normal series.

Discover why every group of order 32 is solvable through a detailed proof using the concept of normal series and Sylow theorems. Explore key group theory concepts like p-groups, centers of groups, and abelian factor groups. Suitable for abstract algebra studies at the undergraduate level.

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