TITLE: Order of a Coset of $U(32)$
QUESTION [1 upvotes]: I suspect I'm bungling something trivial, but I can't see it. I am working on the following question:

Let $G=U(32)$ (that is integers relatively prime less than $32$ under multiplication) and $H=(1, 15)$. The group $G/H$ is isomorphic to
  one of $\mathbf{Z}_8 , \mathbf{Z}_2 \oplus \mathbf{Z}_4, \text{or } \mathbf{Z}_2 \oplus \mathbf{Z}_2 \oplus \mathbf{Z}_2$. Determine which one by
  elimination.

I just checked the coset $3H$:
$$3H = \big(3,13\big) \\ 3^2H = \big(9,3^215\big) = \big(9,7\big)\\ 3^4H = \big(81,3^415\big) = \big(17,31\big)\\ 3^8H = \big(3^8,3^815\big) = \big(1,15\big)$$
I concluded then, that $G/H$ has an element of order $8$ and so must be isomorphic to $\mathbf{Z}_8$. The back of the book says this is not correct though. What am I missing?

REPLY [1 votes]: Your assertation is correct. In fact you can prove that $U(32) \cong \langle 15 \rangle \times \langle 3 \rangle \cong \mathbb{Z}_2 \times \mathbb{Z}_8$. In fact in general $U(2^k) \cong \mathbb{Z}_2 \times \mathbb{Z}_{2^{k-2}}$ whenever $k \ge 2$. Now you have that $H \cong \langle 15 \rangle \times \langle e \rangle$, so
$$U(32)/H \cong \left(\langle 15 \rangle \times \langle 3 \rangle \right)/(\langle 15 \rangle \times \langle e \rangle) \cong (\langle 15 \rangle/\langle 15 \rangle) \times (\langle 3 \rangle/\langle e \rangle) \cong \langle e \rangle \times \langle 3 \rangle \cong \mathbb{Z}_8$$
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