Fix 31.3-2 part 2

docs/Chap31/31.3.md

@@ -17,6 +17,17 @@ By Lagrange's Theorem, any subgroup of $\mathbb Z_9$ must be of order $1, 3, 9$.
 - $\\{3, 6, 9\\} \cong \mathbb Z_3$
 - $\mathbb Z_9$  (not isomorphic to $\mathbb Z_3^2$)
 
+Observe $\mathbb Z_{13}^\*$ consists of units $\\{1, \dots, 12\\}$ and is generated by $2$, i.e. $\mathbb Z_{13}^\* = \langle 2 \rangle$. The generators are thus $\langle 2^n \rangle$ for $n | 12$
+
+- $\langle 2 \rangle$
+- $\langle 2^2 \rangle = \langle 4 \rangle$
+- $\langle 2^3 \rangle = \langle 8 \rangle$
+- $\langle 2^4 \rangle = \langle 3 \rangle$
+- $\langle 2^6 \rangle = \langle -1 \rangle$
+- $\langle 2^{12} \rangle = \langle 1 \rangle$
+
+[Math SE source](https://math.stackexchange.com/a/1352349)
+
 ## 31.3-3
 
 > Prove Theorem 31.14.