I can help analyze the questions in the uploaded image. Let me transcribe and solve these mathematical problems step by step.

Question 1:
In the symmetric group $S_3$, let $H = \{ (1), (12) \}$ and $K = \{ (1), (13) \}$. Show that $HK \neq KH$.

Solution:

• Here, $H$ and $K$ are subgroups of $S_3$.
• Definition: For $HK$, the product of subsets is defined as $HK = \{hk \mid h \in H, k \in K\}$. Similarly, $KH = \{kh \mid k \in K, h \in H\}$.
• We must prove that $HK \neq KH$.

Computation of $HK$:
$H = \{(1), (12)\}, \quad K = \{(1), (13)\}$ $HK = \{h \cdot k \mid h \in H, k \in K\}$ $HK = \{(1) \cdot (1), (1) \cdot (13), (12) \cdot (1), (12) \cdot (13)\} = \{(1), (13), (12), (132)\}$

Computation of $KH$:
$KH = \{k \cdot h \mid k \in K, h \in H\}$ $KH = \{(1) \cdot (1), (1) \cdot (12), (13) \cdot (1), (13) \cdot (12)\} = \{(1), (12), (13), (123)\}$

Clearly, $HK = \{(1), (13), (12), (132)\} \neq KH = \{(1), (12), (13), (123)\}$.
Thus, $HK \neq KH$.

Question 2:
Determine the number of finite subgroups of $(\mathbb{Z}, +)$.

• Solution: The group $(\mathbb{Z}, +)$ is the additive group of integers.
  • It is infinite and cyclic, generated by 1.
  • The only proper subgroup of $(\mathbb{Z}, +)$ is $\{0\}$.
  • Hence, the number of finite subgroups is $\mathbf{0}$.

Answer: $0$ (Option (a)).

Question 3:
Determine the number of subgroups of the multiplicative group $G = \{ a, a^2, \dots, a^{10} = e \}$.

• Solution:
  • $G$ is a cyclic group of order 10.
  • The number of subgroups of a cyclic group of order $n$ is equal to the number of divisors of $n$.
  • The divisors of 10 are $1, 2, 5, 10$.
  • Hence, the number of subgroups is 4.

Answer: $\mathbf{4}$ (Option (c)).

Question 4:
If $H, K$ are proper subgroups of $G$, then:
(a) $H \cup K = G$
(b) $H \cap K = \phi$
(c) $H \cap K = \{e\}$
(d) $H \subseteq K \text{ or } K \subseteq H$.

• Solution:
  • If $H, K$ are proper subgroups, then $H \cap K \neq \phi$ since subgroups always contain the identity element $e$.
  • Therefore, option (b) is incorrect.
  • $H \cap K = \{e\}$ is not guaranteed unless $H$ and $K$ are disjoint, which is not stated.
  • $H \cup K = G$ is also not guaranteed unless $H$ and $K$ cover all elements, which contradicts the fact they are proper.
  • The correct statement is $H \subseteq K \text{ or } K \subseteq H$.

Answer: $\mathbf{(d)}$.

Let me know if you'd like detailed solutions for the remaining questions.

Questions to Explore Further:

1. What are the general properties of symmetric groups $S_n$?
2. How do you calculate the number of subgroups in a cyclic group of any given order $n$?
3. Explain why the group $(\mathbb{Z}, +)$ has no finite subgroups except $\{0\}$.
4. How do we compute the product of subgroups $HK$ and check if $HK = KH$?
5. What is the significance of divisors in determining the subgroups of cyclic groups?

Tip:

When working with cyclic groups, always start by determining the order of the group and its divisors, as this simplifies subgroup calculations.