1
$\begingroup$

Suppose $H_{1}$ and $H_{2}$ are both subgroups of the group G. We want to show $$H_{1}\bigcap H_{2}\leq G$$

This means(by the sub-group test): $$b^{-1} \in H_{1}\bigcap H_{2}$$ and $$ab \in H_{1}\bigcap H_{2}$$ for all $a,b$

The intersection of the 2 subgroup is non-empty since the identity element of is contained in both subgroups and thus their intersection must necessarily contain e.

I proceed next to show that if $$ab$$ is in the intersection then it must necessarily be in both subgroups. This implies from the inverse property of subgroup that $$b^{-1}$$ is in both subgroup.

But does this necessarily means that $$b^{-1}$$ is in the intersection of the 2 subgroups? I'm not inclined to say yes because while it is necessarily true that anything that is in the intersection must be in the 2 subgroup, the converse is certainly not true.

$\endgroup$
1
$\begingroup$

This is how it's usually done: Take $a, b \in H_1\cap H_2$. That means that we have $a, b \in H_1$, and we have $a, b \in H_2$. Since both $H_1$ and $H_2$ are subgroups of $G$, we know that $ab \in H_1$ and $ab \in H_2$. Therefore, $ab \in H_1\cap H_2$.

Showing that $a^{-1} \in H_1\cap H_2$ is very similar: Since $a \in H_1$, we know that $a^{-1} \in H_1$. In the same way we have $a^{-1}\in H_2$. This means exactly that $a^{-1} \in H_1\cap H_2$.

$\endgroup$
  • $\begingroup$ I assume there must exists a bicondition relating the subgroups and the intersection of all the subgroup. This means that for any element, x, in all the subgroups, it must necessarily be contained in the intersection. $\endgroup$ – Mathematicing Jun 2 '15 at 8:00
  • $\begingroup$ @Mathematicing That is what intersection means. For any kind of set $X$ with subsets $Y, Z$, the intersection $Y\cap Z$ is exactly those elements that are in both $Y$ and $Z$. In this case, the intersection $H_1\cap H_2$ is exactly those elements of $G$ that are in both $H_1$ and in $H_2$. $\endgroup$ – Arthur Jun 2 '15 at 8:01
2
$\begingroup$

There are some issues with your reasoning. You want to show that $H_1\cap H_2$ is a subgroup of $G$.

For this you need to show that $b^{-1}\in H_1\cap H_2$ and $ab\in H_1\cap H_2$ for all $a$ and $b$ in $\mathbf H_1\cap \mathbf H_2$

(You didn't write the part in bold above, which is crucial.)

Then you write:

I proceed next to show that if $ab$ is in the intersection then it must necessarily be in both subgroups.

This is a tautology and will lead to nothing. If $ab$ is in the intersection $H_1\cap H_2$ then by definition it is in both $H_1$ and $H_2$. This has nothing to do with group theory.

What you ought to so is show that if $a, b\in H_1\cap H_2$ then $ab\in H_1\cap H_2$. This is true. For this gives $a, b\in H_1, H_2$, and by the subgroup property, we have $ab\in H_1, H_2$, which is same as saying that $ab\in H_1\cap H_2$.

Similarly for the inverse part.

$\endgroup$
  • $\begingroup$ That reasoning was really bad. It was done so in spur. Thanks for pointing it out. $\endgroup$ – Mathematicing Jun 2 '15 at 8:09
0
$\begingroup$

First you need to show that $H_1 \cap H_2$ is non empty , this is true because $e \in H_1$ and $e \in H_2$ because they are both subgroups of $G$ and so $e \in H_1 \cap H_2$

Now you need to show that if $a \in H_1 \cap H_2$ and $b \in H_1 \cap H_2$ then $ab \in H_1 \cap H_2$, this is also true because $a \in H_1 \cap H_2 \implies a \in H_1$ and $a \in H_2$ and the same thing for $b$ and so $ab \in H_1$ and $ab \in H_2$ and so $ab \in H_1 \cap H_2$

Finally you need to show that if $a \in H_1 \cap H_2$ then $a^{-1} \in H_1 \cap H_2$ , this is also true because $a \in H_1 \cap H_2 \implies a \in H_1 \implies a^{-1} \in H_1$ and also $a \in H_1 \cap H_2 \implies a \in H_2 \implies a^{-1} \in H_2$ and so $a^{-1} \in H_1 \cap H_2$ and so the intersection is a subgroup

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.