What is the order of the multiplicative group?

Question

According to Lagrange's Theorem, the multiplicative group $(\mathbb{Z}_{54})^\times$ cannot contain a subgroup of which order:

A: $9$
B: $18$
C: $6$
D: $12$

I think that it is D because $12$ is not a factor of $54$.

Answer 1

By the Euler totient function, we know that the order of the multiplicative group of $ \mathbb{Z}_{n}=\phi (n)$ Which is given by the formula

$\phi(n)=n\prod_{p | n} \frac{p-1}{p}$

Hence

$\phi(54)=54\frac{3-1}{3} \frac{2-1}{2} =18 $

By Lagrange's theorem, the order of the subgroup must divide the order of the group since 12 does not divide 18 The answer is D

Answer 2

Hint: The multiplicative group of $\Bbb Z_{54}$ has 18 elements. Just compute the Eulerian phi function.

Answer 3

Since the underlying set of $G:=(\Bbb Z_{54})^\times$ is (equivalent to)

$$\{a\in\Bbb N\mid \gcd(a,54)=1\land a\in\{0,1,2,\dots, 53\}\},$$

by definition, its order is $\varphi (54)=2^0(2-1)3^2(3-1)=18$, where $\varphi $ is Euler's totient function.

The divisors of $18$ are $1,2,3,6,9,18$, so, by Lagrange's Theorem, $12$ cannot be the order of a subgroup of $G$.

Hence the answer is indeed D, but not for the reason you stated, which, as pointed out in the comments, is flawed.