Prove that $G = \{ a + b (\sqrt{ 2 })\mid a, b \in \mathbb{Q} , a,b \text{ not simultaneously zero}\}$ is a subgroup of $\mathbb{R}^*$ (nonzero real numbers) under the group operation of multiplication
Sketch of Solution
We notice that $1 + 0 \sqrt{ 2 } \in G$ shows non-empty, so we may proceed by the subgroup criterion. It suffices to show that, for $a+b \sqrt{ 2 }, c+d\sqrt{ 2 } \in G$, $$(a+b\sqrt{ 2 }) \cdot (c + d\sqrt{ 2 })^{-1} \in G$$ Notice that we can rewrite $$(a+b\sqrt{ 2 }) \cdot (c + d\sqrt{ 2 })^{-1} = \frac{a+b\sqrt{ 2 }}{c + d \sqrt{ 2 }} = \frac{a+b\sqrt{ 2 }}{c + d \sqrt{ 2 }} \cdot \frac{c-d\sqrt{ 2 }}{c - d \sqrt{ 2 }} = (ac-2bd) + \sqrt{ 2 }(bc-ad)$$ which is clearly an element of $G$, as needed.
List every generator of each subgroup of order $8$ in $Z_{32}$
Sketch of Solution
We're looking for elements $n \in Z_{32}$ such that $\gcd(n,32) = 4$ for which we can test each of $\left \{ 0,4,8,12,16,20,24,28 \right \}$. Our final answer is: $\left \{ 4,12,20,28 \right \}$.
Show that $A_{10}$ contains an element of order $15$. That includes showing that the permutation you suggest is actually in $A_{10}$
Sketch of Solution
Recall that the order of a permutation is the $\mathrm{lcm}$ of the cycle-type. Thus an element of order $15$ can be of cycle type $(3,5)$. Consider:
$$(1 2 3 4 5) (678) = (68)(67)(15)(14)(13)(12)$$which has positive sign since the number of permutations is even.
Suppose that $g^n = e.$ Show that the order of $g$ divides $n$.
Sketch of Solution
First note that the order must be less than $n$, and thus finite. The order is also a positive number by definition, so the set $$S = \left \{ k \in \mathbb{N} \mid g^k = e \right \}$$ is a bounded non-empty (for example, $n \in S$) set of integers. By the Well-Ordering Principle, there exists $m\in S$ minimal. Using the Division Algorithm we can write: $n = qm + r$, where $q,r$ are non-negative integers, $0 \leq r < m$. We'll argue that $r = 0$ by seeing: $$e = g^n = g^{qm + r} = (g^m)^q g^r = e \cdot g^r$$ If $r \neq 0$ this contradicts minimality of $m$.