Question

Let \( G \) be a group satisfying the property that \( f: G \to \mathbb{Z}_{221} \) is a homomorphism implies \( f(g) = 0 \), \( \forall g \in G \). Then a possible group \( G \) is

• A. \( \mathbb{Z}_{21} \)
• B. \( \mathbb{Z}_{51} \)
• C. \( \mathbb{Z}_{91} \)
• D. \( \mathbb{Z}_{119} \)

Hint:

When analyzing homomorphisms, check the orders of the groups involved to determine possible values for the group \( G \).

Answer 1

The Correct Option is A

Step 1: Factor $221$

$$221 = 13 \times 17$$

So $\mathbb{Z}_{221} \cong \mathbb{Z}_{13} \times \mathbb{Z}_{17}$ (since $\gcd(13, 17) = 1$).

Step 2: Understanding homomorphisms from $G$ to $\mathbb{Z}_{221}$

For a homomorphism $f: G \to \mathbb{Z}_{221}$:

• By the first isomorphism theorem, $\text{Im}(f)$ is a subgroup of $\mathbb{Z}_{221}$
• The subgroups of $\mathbb{Z}_{221}$ are cyclic groups of orders dividing $221$
• The divisors of $221$ are: $1, 13, 17, 221$

Step 3: Condition for trivial homomorphism

For every homomorphism $f: G \to \mathbb{Z}_{221}$ to be trivial, we need:

• $G$ has no quotient group isomorphic to any non-trivial subgroup of $\mathbb{Z}_{221}$
• Equivalently, $G$ has no quotient of order $13$, $17$, or $221$

By the fundamental theorem of homomorphisms, if $f: G \to H$ is a surjective homomorphism, then $H \cong G/\ker(f)$.

Step 4: Check each option

(A) $\mathbb{Z}_{21}$:

$21 = 3 \times 7$

For a homomorphism $f: \mathbb{Z}{21} \to \mathbb{Z}{221}$:

• $f$ is determined by $f(1)$
• We need $21 \cdot f(1) = 0$ in $\mathbb{Z}_{221}$
• This means the order of $f(1)$ divides $\gcd(21, 221) = 1$

Therefore, $f(1) = 0$, so $f$ is trivial. 

(B) $\mathbb{Z}_{51}$:

$51 = 3 \times 17$

For $f: \mathbb{Z}{51} \to \mathbb{Z}{221}$:

• We need $51 \cdot f(1) = 0$ in $\mathbb{Z}_{221}$
• The order of $f(1)$ divides $\gcd(51, 221) = 17$

We can have $f(1)$ be an element of order $17$ in $\mathbb{Z}_{221}$. For example, $f(1) = 13$ has order $17$.

Check: $51 \cdot 13 = 663 = 3 \times 221 \equiv 0 \pmod{221}$ 

This gives a non-trivial homomorphism. 

(C) $\mathbb{Z}_{91}$:

$91 = 7 \times 13$

For $f: \mathbb{Z}{91} \to \mathbb{Z}{221}$:

• $\gcd(91, 221) = 13$

We can construct a non-trivial homomorphism with image of order $13$. 

(D) $\mathbb{Z}_{119}$:

$119 = 7 \times 17$

For $f: \mathbb{Z}{119} \to \mathbb{Z}{221}$:

• $\gcd(119, 221) = 17$

We can construct a non-trivial homomorphism with image of order $17$. 

Answer: (A)