# [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.

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I think the authors are "Dummit and Foote"; not "Dummit and Forte" – Amitesh Datta Jun 14 '11 at 6:16
by definition is OK. – wxu Jun 14 '11 at 6:17
Mathematical induction would not work even for a countably infinite collection. The collection is supposed to be "arbitrary", not assumed finite or countable. The argument is more direct than induction. – Jonas Meyer Jun 14 '11 at 6:19
Thanks a lot for the help guys. – user12091 Jun 14 '11 at 7:03
Well, you could prove it by transfinite induction, I suppose. (Not that you should or that it is at all necessary or helpful to do it that way...) – Pete L. Clark Jun 14 '11 at 8:41

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 $G$?

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Dear Amitesh, A minor point, but in your steps (1) and (2) you write the group operation additively, rather than multiplicatively (as would seem more natural for a question that doesn't require the groups to be abelian). Is there a reason for this? Best wishes, – Matt E Jun 14 '11 at 7:00
Dear Matt E, I have been thinking too much about commutative rings lately! Thank you for noting this. I will fix it. Regards, Amitesh – Amitesh Datta Jun 14 '11 at 7:04
I should mention that exercise 3 is a beautiful exercise because it looks (at a first glance, at least) like a simple exercise in logic. However, it is more subtle than it first appears. In particular, it is a good (challenging) exercise to first ask a student who has only seen the definition of a group to solve. Once the student has seen Exercises 1 and 2 above, it is then again worth asking him to solve Exercise 3 (but without explicitly mentioning Exercises 1 and 2 as hints). The student will then appreciate how the notion of a "subgroup" provides a more unified approach to group theory. – Amitesh Datta Jun 14 '11 at 11:17
@Amitesh Datta Please see my answer below: – user38268 Aug 1 '11 at 12:39
@DLim Thank you for noting this; I have seen and commented on your answer below. – Amitesh Datta Aug 2 '11 at 0:49

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!

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+1 The solutions are certainly correct. However, there is an easier way to solve Exercise 2. Indeed, if $z\in \textbf{Z}(G)$, then $zx=xz$ for all $x\in G$. Let us invert both sides of the equation $zx=xz$ and obtain $x^{-1}z^{-1}=(zx)^{-1}=(xz)^{-1}=z^{-1}x^{-1}$. Since every element in $G$ is of the form $x^{-1}$ for some $x\in G$, it follows that $z^{-1}\in \textbf{Z}(G)$. – Amitesh Datta Aug 2 '11 at 0:38
Also, the proof you have given in your first paragraph extends to arbitrary (not necessarily finite) intersections of subgroups. – Amitesh Datta Aug 2 '11 at 0:43
@Amitesh Datta Thanks for checking them. I won't mention the proof here but to prove exercise 1 is just an exercise in logic. – user38268 Aug 2 '11 at 12:48
@Nice proof by the way. Just using the fact that $xz = zx$! I must be blind or something... – user38268 Aug 2 '11 at 12:52