Question

Which ONE of the following statements is FALSE?

• A. There is a surjective group homomorphism from the additive group of rational numbers to the multiplicative group of all complex roots of unity.
• B. The multiplicative group of complex numbers of modulus one is isomorphic to a quotient group of the additive group of real numbers.
• C. Any group homomorphism from the multiplicative group of nonzero complex numbers into the group of all invertible \( 2 \times 2 \) matrices with real entries has a nontrivial kernel.
• D. There exists a group homomorphism from the symmetric group on \( n \) symbols into the multiplicative group of all invertible \( n \times n \) matrices with real entries, which has a trivial kernel.

Hint:

When checking group homomorphisms, the kernel is important. A nontrivial kernel implies that some elements of the domain map to the identity in the codomain, which can often be verified through the properties of the groups involved.

Answer 1

The Correct Option is C

Step 1: Read the question.
We must find the one statement that is not true. So we test each option and stop when one breaks.

Step 2: Check option A.
All complex roots of unity together form the group $\mathbb{Q}/\mathbb{Z}$. The natural quotient map sends $\mathbb{Q}$ onto $\mathbb{Q}/\mathbb{Z}$, so a surjective map from the rationals does exist. This option is true.

Step 3: Check option B.
The circle group of unit modulus numbers is $\{e^{2\pi i t}\}$. The map $t \mapsto e^{2\pi i t}$ takes $\mathbb{R}$ onto it with kernel $\mathbb{Z}$, so the circle is $\mathbb{R}/\mathbb{Z}$. This option is true.

Step 4: Check option C.
Take the map $a+bi \mapsto \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$. This sends nonzero complex numbers into invertible $2\times 2$ real matrices, and the only number going to the identity is $1$. So the kernel is trivial here. That means the claim 'every such map has a nontrivial kernel' is wrong.

Step 5: Conclusion.
Option C is the false one, since we built a map with trivial kernel.
\[ \boxed{\text{Option C is FALSE}} \]