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If A and B are subgroups of G then is A∩B and A∪B also subgroups of G? I think that they both should be because they are both:

  1. Closed: A∩B and A∪B are both in G.

  2. Associative: A∩B and A∪B inherit this ability from G.

  3. Have the identity element: A∩B are all the elements A and B share and A and B both have the identity element, thus, A∩B has the identity element. A∪B are all the elements A or B have and both A and B have the identity, thus A∪B has the identity element.

  4. Have an inverse for each element: A∩B are all the elements A and B share, thus, if A and B share an element they must also share its inverse since A and B are subgroups. A∪B are all the elements A or B have so all elements will have an inverse.

Is my logic correct?

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  • $\begingroup$ How do you show that $A\cup B$ is closed? $\endgroup$
    – user276115
    Commented Sep 29, 2016 at 15:34
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    $\begingroup$ Consider $G=\mathbb{Z}$ and its subgroups $A=2\mathbb{Z}$ and $B=3\mathbb{Z}$. Then $A\cup B$ is not closed since $2+3=5\not\in A\cup B$. $\endgroup$
    – Olivier Moschetta
    Commented Sep 29, 2016 at 15:43
  • $\begingroup$ The binary operation of $2\mathbb{Z}$ is $2^{n}$ and the binary operation of $3\mathbb{Z}$ is $3^{n}$ so how can the binary operation of $A\cup B$ be addition? I'm just confused how you can add to elements in $A\cup B$. $\endgroup$
    – PiccolMan
    Commented Sep 29, 2016 at 15:55

1 Answer 1

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$A \cup B$ is never a subgroup unless it is either $A$ or $B$ (i.e $A \subseteq B$ or $B \subseteq A$). Otherwise there are $a \in A$ and $b \in B$ such that $a \notin B$ and $b \notin A$. $ab \notin A$, otherwise $b = a^{-1}(ab)$ would be in $A$, and $ab \notin B$, otherwise $a = (ab)b^{-1}$ would be in $B$.

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  • $\begingroup$ Thanks for explaining. Are you saying that $A\cup B$ is not closed? $\endgroup$
    – PiccolMan
    Commented Sep 29, 2016 at 16:11
  • $\begingroup$ @PiccolMan Yes they are! It should seem unlikely for the union to be closed, because adding something from A to something from B has no reason to live anywhere in particular. $\endgroup$
    – Stella Biderman
    Commented Sep 29, 2016 at 16:40

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