Question

For a positive integer n, let $U (n) = {\bar{r} \in Z_{n} : g c d (r, n) = 1}$ be the group under multiplication modulo n. Then, which one of the following statements is TRUE?

Answer: 𝑈(8) is isomorphic to 𝑈(12)

isomorphic Symbol → $\approx$
not isomorphic Symbol → $\cancel{\approx}$

Option (1)

U (5) \approx Z_{5 - 1} \approx Z_{4}, and

U (8) = U (2^{3}) \approx Z_{2} \times Z_{2^{3 - 2}} \approx Z_{2} \times Z_{2} \cdot

Since, $Z_4\; \cancel{\approx}\; Z_2\times Z_2$
Hence, $U(5)\; \cancel{\approx}\; U(8)\cdot$
So, option (1) is wrong.

Option (2)

U (10) = U (2 \times 5) \approx U (2) \times U (5) \approx Z_{1} \times Z_{5 - 1} \approx Z_{1} \times Z_{4}, and

U (12) = U (3 \times 4) \approx U (3) \times U (4) \approx Z_{2} \times Z_{2} \cdot

Since, $Z_1\times Z_4\; \cancel{\approx}\; Z_2\times Z_2$
Hence, $U(10)\; \cancel{\approx}\; U(12)\cdot$
So, option (2) is also wrong.

Option (3)

U (8) = U (2^{3}) \approx Z_{2} \times Z_{2^{3 - 2}} \approx Z_{2} \times Z_{2}, and

U (10) = U (2 \times 5) \approx U (2) \times U (5) \approx Z_{1} \times Z_{5 - 1} \approx Z_{1} \times Z_{4} \cdot

Since, $Z_2\times Z_2\; \cancel{\approx}\; Z_1\times Z_4$
Hence, $U(8)\; \cancel{\approx}\; U(10)\cdot$
So, option (3) is also wrong.

Option (4)

U (8) = U (2^{3}) \approx Z_{2} \times Z_{2^{3 - 2}} \approx Z_{2} \times Z_{2}, and

U (12) = U (3 \times 4) \approx U (3) \times U (4) \approx Z_{2} \times Z_{2} \cdot

Since, $Z_{2} \times Z_{2} \approx Z_{2} \times Z_{2}$
Hence, $U (10) \approx U (12) \cdot$
So, option (4) is correct.

So, the correct answer is option-4, $\to$ 𝑈(8) is isomorphic to 𝑈(12)