# If $P \leq G$, $Q\leq G$, are $P\cap Q$ and $P\cup Q$ subgroups of $G$? [closed]

$P$ and $Q$ are subgroups of a group $G$. How can we prove that $P\cap Q$ is a subgroup of $G$? Is $P \cup Q$ a subgroup of $G$?

Reference: Fraleigh p. 59 Question 5.54 in A First Course in Abstract Algebra.

• The proof that $P\cap Q$ is a subgroup of $G$ is awfully straightforward; where are you stuck? For the second question, what if $G=\Bbb Z$, $P=2\Bbb Z$ is the set of multiples of $2$, and $Q=3\Bbb Z$ is the set of multiples of $3$? Dec 7, 2012 at 22:48
• I am sorry, I am not familiar with the proof of $P \cap Q$ and that it is a subgroup of $G$. Dec 7, 2012 at 22:52
• You’re expected to come up with a proof that $P\cap Q$ is a subgroup of $G$, not to regurgitate one that you’ve seen before. Can you write down what it means to say that $H$ is a subgroup of $G$? Dec 7, 2012 at 22:56
• This homework should and can be done by everyone who wants to learn group theory, especially since you don't need any ideas. Just play around with the new notions. You won't learn it by just asking others and copying the solution. But of course nobody cares what I write here :) Jan 27, 2013 at 22:51
• @MartinBrandenburg I care! And your point is good. When explicitly tagged homeowork, and given a problem of this nature, I get tired of folks spitting out full-fledged solutions. Jan 27, 2013 at 22:55

$P$ and $Q$ are subgroups of a group $G$. Prove that $P \cap Q$ is a subgroup.

Hint 1:
You know that $P$ and $Q$ are subgroups of $G$. That means they each contain the identity element, say $e$ of $G$. So what can you conclude about $P\cap Q$? If $e \in P$ and $e \in Q$? (Just unpack that means for their intersection.)

Hint 2:
You know that $P, Q$ are subgroups of $G$. So they are both closed under the group operation of $G$. If $a, b \in P\cap Q$, then $a, b \in P$ and $a, b \in Q$. So what can you conclude about $ab$ with respect to $P\cap Q$? This is about proving closure under the group operation of $G$.

Hint 3:
You can use similar arguments to show that for any element $c \in P\cap Q$, $c^{-1} \in P\cap Q$. That will establish that $P\cap Q$ is closed under inverses.

Once you've completed each step above, what can you conclude about $P\cap Q$ in $G$?

$P$ and $Q$ are subgroups of a group $G$. Is $P\cup Q$ a subgroup of $G\;$?

Here, you need to provide only one counterexample to show that it is not necessarily the case that $P\cup Q$ is a subgroup of $G$.

• Suppose, for example, that your group $G = \mathbb{Z}$, under addition.

Then we know that $P = 2\mathbb{Z} \le \mathbb{Z}$ under addition (all even integers), and $Q = 5\mathbb{Z} \le \mathbb{Z}$ under addition (all integer multiples of $5$).

So $P \cup Q$ contains $2\in P,$ and $5 \in Q.\;\;$ But:$\;$ is $\;2 + 5 = 7 \in P\cup Q\;$?

So what does this tell regarding whether or not $P \cup Q$ is a subgroup of $\mathbb{Z}\;$?
• Good answer. For the sake of completeness, if $P\cup Q$ is a subgroup what does it implies?
– leo
Dec 8, 2012 at 23:32
• @leo $P\cup Q$ is a subgroup if and only if $P\subseteq Q$ or $Q\subseteq P$.
– Pedro
Apr 6, 2013 at 1:28
• @PeterTamaroff I know :-) It was a rethorical question to point out that it is important to note that the only scenario in which $P\cup Q$ is a subgroup is the one you've just described.
– leo
Apr 6, 2013 at 5:19
• @leo I know you know. I was just "reflex responding".
– Pedro
Apr 6, 2013 at 5:22
• @amWhy: this deserves a badge! +1 May 3, 2013 at 0:36

$P\cup Q$ need not be a subgroup. For example, $2\mathbb Z$ and $3\mathbb Z$ are subgroups of $\mathbb Z$, the group of integers under ordinary addition, but their union is not a subgroup because $2\in 2\mathbb Z$, $3\in 3\mathbb Z$ but $2+3=5\not\in 2\mathbb Z\cup 3\mathbb Z$.

To show that $P\cap Q$ is a subgroup, note that $e\in P$ and $e\in Q$ so that $e\in P\cap Q$ and $P\cap Q$ is nonempty. If $a\in P\cap Q$ and $b\in P\cap Q$, then $a\in P$, $a\in Q$, $b\in P$ and $b\in Q$, so $ab^{-1}\in P$ and $ab^{-1}\in Q$, so $ab^{-1}\in P\cap Q$.

• What about closure for $P\cap Q$? Dec 7, 2012 at 22:59
• Nearly any example one tries will result in $P \cup Q$ not being a subgroup (one would need $P \subseteq Q$ or vice versa to avoid this). All of this suggests a disappointing lack of effort on OP's part. Dec 8, 2012 at 1:03

For the last part: $P \cup Q$ is a subgroup if and only if $P \subset Q$ or $Q \subset P$.

$\Leftarrow$ is obvious

$\Rightarrow$ Assume by contradiction that this is not true. Pick $x \in P \backslash Q$ and $y \in Q \backslash P$.

Then $x,y \in P \cup Q$ implies $x+y \in P \cup Q$, hence $x+y$ in either $P$ or $Q$. But then, either $x=(x+y)-y \in Q$ or $y=(a+y)-x \in P$ contradiction.

Since $1 \in H_{1} \cap H_{2}$, the intersection is not empty. Now let $x, y \in H_{1} \cap H_{2}$. Then $xy^{-1} \in H_{1}$ and $H_{2}$ since $x, y \in H_{1}$ and $H_{2}$. Thus $H_{1} \cap H_{2} \leqslant G$.

Consider $H_{1} = 2\mathbb{Z}$ and $H_{2} = 3\mathbb{Z}$. Then we have $H_{1} \cup H_{2} = 2\mathbb{Z} \cup 3\mathbb{Z}$, which does not have $3 - 2 = 1$.

• @GitGud Then it is good time for OP to learn it! It is not to hard to see that other axioms can be inherited. Jan 27, 2013 at 22:34
• I decided to delete my comment because I'd have to say the same to every answer so far. Jan 27, 2013 at 22:37
• Thank you so much Sir....thanking you for you support and guidance Jan 27, 2013 at 23:42
• @DenishSen No problem. Actually, the intersection of a (nonempty) collection of arbitrary subgroups is a subgroup, so intersection is a nice way to define a subgroup generated by a subset of a given whole group. So far the only time when I think taking union is nice is when we have an increasing chain because we can put $x, y$ into one subgroup and see that $xy^{-1}$ is in that group (and therefore in the union). Jan 28, 2013 at 1:56

Counter: Letting $x = \left( \textrm{1 2 3} \right), y = \left( \textrm{1 2} \right)$. Then $\langle x \rangle , \langle y \rangle$ are subgroups of $S_3$, but their union is not (it is not closed).

Intersection: if $x, y \in H \cap K$, then $x, y \in H, K$. From this, closedness, existence of inverses and identity in $H, K$ follow easily, and so in $H \cap K$.

• Thank you Sir for your guidance and interest in solving the exercise. Jan 27, 2013 at 23:52