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Consider the following statements.

I. The group $(\mathbb{Q}, +)$ has no proper subgroup of finite index.

II. The group $(\mathbb{C} \setminus {0}, \cdot)$ has no proper subgroup of finite index.

Which one of the following statements is true?

(A) Both I and II are TRUE.

(B) I is TRUE but II is FALSE.

(C) II is TRUE but I is FALSE.

(D) Neither I nor II is TRUE.
in MA2021 by Professor | 151 views

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The correct answer is:

(B) I is TRUE but II is FALSE.


Explanation:

I. The group $(\mathbb{Q}, +)$ has no proper subgroup of finite index:

The group $(\mathbb{Q}, +)$ of rational numbers under addition is a divisible abelian group. Divisibility means that for any element $x$ in the group and any positive integer $n$, there exists an element $y$ in the group such that $ny = x$. In the case of $(\mathbb{Q}, +)$, for any rational number $x$ and any positive integer $n$, we can always find a rational number $y$ such that $ny = x$. This property implies that every subgroup of $(\mathbb{Q}, +)$ has infinite index, meaning that there are no proper subgroups of finite index in $(\mathbb{Q}, +)$. Thus, statement I is TRUE.


II. The group $(\mathbb{C} \setminus {0}, \cdot)$ has no proper subgroup of finite index:

To show that statement II is false, we need to demonstrate the existence of a proper subgroup of finite index in the group $(\mathbb{C} \setminus {0}, \cdot)$.

Consider the subgroup generated by the complex number $i$, which is denoted as $\langle i \rangle$. This subgroup consists of all integer powers of $i$: $\{i^k : k \in \mathbb{Z}\}$. Clearly$\langle i \rangle$ is a proper subgroup of $(\mathbb{C} \setminus {0}, \cdot)$ of order two.

Therefore, statement II is FALSE, and the correct answer remains (B) I is TRUE but II is FALSE.


Extra: 

Let's clarify how divisibility implies the non-existence of proper subgroups.

 

Consider a group $G$ that is divisible. For any element $g$ in $G$ and any positive integer $n$, there exists an element $h$ in $G$ such that $nh = g$. In other words, every element of $G$ can be written as a multiple of any positive integer.

Now, suppose $H$ is a nontrivial proper subgroup of $G$. Since $H$ is nontrivial, it contains at least one element other than the identity element. Let $h \in H$ be such an element.

Because $G$ is divisible, for any positive integer $n$, there exists an element $x$ in $G$ such that $nx = h$. But if $nx \in H$, then $h \in H$ for all positive integers $n$. This means that $H$ contains all multiples of $h$.

Since $H$ contains all multiples of $h$, it contains all elements of the form $nh$ for all positive integers $n$. Since $H$ contains $h$, it must also contain $2h$, $3h$, $4h$, and so on. In other words, $H$ contains an infinite number of distinct elements, making it infinite.

Since $H$ is an infinite proper subgroup of $G$, $G$ cannot have any proper subgroups of finite index. Any proper subgroup of $G$ would be either trivial (having just the identity element) or infinite.

So, if $G$ is a divisible group, it cannot have any proper subgroups of finite index.

by Professor
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