Subgroups of groups of order $2^{a-1}$

The context here is the following exercise

Let $m=2^a$ with $a > 2$. Show that $\mathbb{Q}(\theta_m)$ contains exactly three quadratic subfields.

By Galois theory, this reduces to the problem of showing that that the multiplicative group $(\mathbb{Z}/2^{a})^\times$ has exactly three subgroups of index 2.

However, this is as far as I get. I have found one subgroup of index 2, namely all $x \equiv 1 \pmod{4}$, but this seems to be the only such subgroup.

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The multiplicative group $(\mathbb{Z}/2^{a})^\times$ is isomorphic to $C_2 \times C_{2^{a-2}}$. – lhf Sep 28 '11 at 11:42

Hint: $(\mathbf{Z}/{2^a\mathbf{Z}})^*$ is isomorphic to $C_{2^{a-2}}\times C_2$. One factor is generated by the coset of $-1$, the other by the coset of $5$. The structure of $(\mathbf{Z}/{2^a\mathbf{Z}})^*$ is derived in many a textbook. If you have problems believing this, then you should take a look at this answer by lhf. You need to be careful and keep track of the exact power of $2$ that divides $5^{2^k}-1$. The factorization given there will help you a lot.
Hint #2: If $[G:H]=2$, then $g^2\in H$ for all $g\in G$.
Thanks! I managed to prove that $(\mathbb{Z}/2^a)^\times \approx C_{2^{a-2}} \times C_2$. So I need to find three subgroups of order $2^{a-2}$ in $C_{2^{a-2}} \times C_2$ and show that there are no more. The subgroup generated by $(1,0)$ and the subgroup generated by $(2,1)$ are two such subgroups. However, there does not seem to be any more. – Fredrik Meyer Sep 28 '11 at 12:55
@Fredrik: The element $(2,1)$ is of order $2^{a-3}$, so the subgroup it generates has index 4. Check that! – Jyrki Lahtonen Sep 28 '11 at 13:06
Ah. After some experimenting with small $a$, it seems obvious :) The three subgroups are the cyclic group generated by $(1,0)$, the cyclic group $(1,1)$, and the non-cyclic group generated by $(2,0)$ and $(0,1)$. – Fredrik Meyer Sep 28 '11 at 18:17