[Dummit and Foote : Algebra] 2.1 Subgroups Question 10(b) Some help/hint please So, the question goes like this "Prove that the intersection of an arbitrary nonempty collection of subgroups of $G$ is again a subgroup of $G$ (do not assume the collection is countable)."
At first, I thought of approaching this question with a normal mathematical induction. ( Assume true for base case, prove that if it's true for $n$ number of set, then it must be true for $n+1$ number of set and so on so forth) However, the bold part of the question confused me. Isn't it the case that if we use MI to prove this, the set is countable? ( although, it could be countably infinite ). Or, does the question actually ask us not to assume the number of sets is finite instead?
Thanks in advance for the pointers. 
 A: The following steps lead to a solution:
(1) Let $\{H_{\alpha}\}_{\alpha\in A}$ be a collection of subgroups of a group $G$ where $A$ is an index set and let $H=\bigcap_{\alpha\in A} H_{\alpha}$ be their intersection. Prove directly from the definitions that $H$ is a subgroup of $G$. For example, if $x\in H$, then by definition of "intersection", $x\in H_{\alpha}$ for all $\alpha\in A$. Since $H_{\alpha}$ is a subgroup of $G$, we know that $x^{-1}\in H_{\alpha}$ for all $\alpha\in A$. In particular, $x^{-1}\in H$ since $H=\bigcap_{\alpha\in A} H_{\alpha}$. 
(2) Prove similarly (and carefully) that if $x,y\in H$, then $xy\in H$. (Hint: think carefully about the definition of "intersection" and relate to the proof given in (1) above.) 
The following exercises will strengthen your understanding of the concepts relevant to the the question you asked:
Exercise 1: Let $H,K\subseteq G$ be subgroups and supppose $H\cup K$ is also a subgroup of $G$. Prove that either $H\subseteq K$ or $K\subseteq H$. (Hint: prove this by contradiction. In particular, if neither of the alternatives in the second sentence holds, choose $x\in H$, $x\not\in K$ and $y\in K$, $y\not\in H$. Note that $x,y\in H\cup K$ and use the fact that $H\cup K$ is a subgroup of $G$.)
Exercise 2: Let $G$ be a group and let $x\in G$. The centralizer of $x$ in $G$ is the set $\textbf{C}_G(x)=\{g\in G:xg=gx\}$. Prove that $\textbf{C}_G(x)$ is a subgroup of $G$.
Exercise 3: Let $G$ be a group with the property that there do not exist three elements $x,y,z\in G$ no two of which commute. Prove that $G$ is abelian. (Hint: fix $x,y\in G$; in order to prove that $x$ and $y$ commute, write $G$ as the union of two appropriately chosen subgroups and appeal to Exercise 1 above. Exercise 2 is relevant.)
Exercise 4: Let $H$ and $K$ be subgroups of a group $G$ such that $HK=KH$. (Let me recall that $HK=\{hk:h\in H \text{ and } k\in K\}$ and $KH=\{kh:h\in H \text{ and } k\in K\}$.) Prove that $HK=KH$ is a subgroup of $G$. Conversely, if $HK$ (or $KH$) is a subgroup of $G$, prove that $HK=KH$. (Hint: the proof (like the solution to your question) is a simple manipulation of the definitions which you should carefully work through.)
Exercise 5: Let $H$ be a subset of $G$ such that for every proper subgroup $K$ of $G$, we have that $H\cap K$ is a subgroup of $G$. Is it necessarily true that $H$ is a subgroup of $G$? 
A: For statement 2 above, choose $x,y$ such that $x,y \in A\cap B$. Then this is the same as saying that ($x$ is in $A$ and $B$) and ($y$ is in $A$ and $B$). Because the logical connective $^$ is associative and commutative we can move the paranthese around and say that this is equivalent to ($x \in A$ and $y \in A$) and ($x \in B$ and $y \in B$).
Because $A$ and $B$ by themselves are subgroups, it follows that $xy \in A$ and $xy \in B$, so that $xy \in A \cap B$.
Of course this proof can be extended to the case of an arbitrary (finite?) intersection of subgroups.
For exercise 2, I can prove a similar problem: Prove that the set $Z(G) = \{z\in G : zx = xz \quad \forall x \in G\}$, i.e. the center of the group is a subgroup of $G$.
Clearly $1_G$ is in $Z(G)$ as $1_Gx = x1_G$ for all $x \in G$. Now suppose $z_1$ and $z_2$ are in $Z(G)$. Then $z_2z_1x = z_2xz_1 = xz_2z_1$ so that $z_1z_2 \in G$. Finally, suppose $z \in Z(G)$. We need to show that $z^{-1} \in G$. If
$\begin{eqnarray*} 
xz &=& zx, \\ \implies z^{-1}(xz)x^{-1} 
&=& z^{-1}(zx)x^{-1} \\ 
&=& (z^{-1}z)xx^{-1} \\
&=& z(z^{-1}x)x^{-1}. \end{eqnarray*}$
Now $z^{-1}(xz)x^{-1} = z(xz^{-1})x^{-1}$ using the fact that $xz = zx$. We conclude that 
$\begin{eqnarray*}z(xz^{-1})x^{-1} &=& z(z^{-1}x)x^{-1}\\
\implies xz^{-1} &=& z^{-1}x. \end{eqnarray*}$
So $z^{-1} \in Z(g)$ and by the facts above, we conclude that $Z(G) \leq G.$
P.S. By the way exercise 1 is part of our recent assignment in algebra!
