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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 $C_2 \times C_{2^{a-2}}$.2011-09-28

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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 $g\in G$.

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    Correct. Well done, @Fredrik2011-09-28