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In more detail, if $G$ is a group and $H_1$, $H_2$ are subgroups of G then $H_1 \cap H_2$ is a subgroup of G. Next, give an example of a particular group $G$ (any one you like), and two different subgroups $H_1$, $H_2$ of $G$ , compute the intersection $H_1 \cap H_2$ , and verify it is indeed a subgroup.

Finally, give three examples showing that $H_1 \cup H_2$ need not be a subgroup of $G$ .

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    $\begingroup$ Check it is just straightforward with the very definition of a subgroup. For exemples, think at the simplest groups you know, it always walks... $\endgroup$ – Léo Mar 6 '14 at 20:39
  • $\begingroup$ thank you very much and i will use those examples :D $\endgroup$ – user131057 May 8 '14 at 0:23
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Theorem Let $G$ be a non-trivial finite group. Then the following are equivalent.

(a) For each pair of subgroups $H_1$ and $H_2$ of $G$, $H_1 \cup H_2$ is a subgroup
(b) $G$ is cyclic of prime-power order.

Proof (b) $\Rightarrow$ (a) follows from the fact that in a cyclic group there is a unique subgroup of order $d$ for each divisor $d$ of $|G|$. (If $G$ has order $p^n$ and $G=\langle g \rangle$, then define $H_i=\langle g^{p^{n-i}} \rangle$. Then $|H_i|=p^i$ and the series $\{1\} =H_0 \subset H_1 \dots \subset H_{n-1} \subset G=H_n$ are all the subgroups of $G$.)
(a) $\Rightarrow$ (b). Assume that $G$ is not cyclic. Then we can find a non-identity $x \in G$, with $\langle x \rangle \subsetneq G$, and take $|\langle x \rangle|$ as large as possible. Also, we can find a $y \in G-\langle x \rangle$. The assumption implies $H=\langle x \rangle \cup \langle y \rangle$ is a subgroup. Clearly $x$, $y$ $\in H$, but $xy \notin H$, since $x$ is not a power of $y$ and vice versa: if $xy=x^i$ for some integer $i$, then $y=x^{i-1} \in \langle x \rangle$, a contradiction. If $xy=y^j$ for some integer $j$, then $x=y^{j-1} \in \langle y \rangle$, whence $\langle x \rangle \subseteq \langle y \rangle$, and hence $\langle x \rangle = \langle y \rangle$, by the maximality of $\langle x \rangle$. Again a contradiction. So $G$ must be cyclic. If $|G|$ has two different prime factors $p$ and $q$, the by Cauchy's Theorem we can find elements $a$ and $b$ of that orders respectively, and $a$ and $b$ are powers of a $g$ with $\langle g \rangle = G$. Again, $\langle a \rangle\cup \langle b \rangle$ is a subgroup of $G$, but it does not contain $ab$, that has order $pq$. A contradiction, the order of $G$ must be divisible by a single prime, whence $G$ is of prime-power order.$\square$

One can also use the well-known fact that a group can never be the union of two proper subgroups. Anyway, the theorem provides you a slew of examples for the second part of your question.

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  • $\begingroup$ very nice big help and this is very great help for me :D $\endgroup$ – user131057 May 8 '14 at 0:22
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Use the definition of a subgroup. You know that they both share the identity element, and for any element they share, they must also share the inverse. Can you take it from here?

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  • $\begingroup$ yes thank you for your help :D $\endgroup$ – user131057 May 8 '14 at 0:22
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Let $G=S_3=\{(1),(1,2),(1,3),(2,3),(1,2,3),(1,3,2) \}$ and $$H_1=\{(1),(1,3) \}$$ $$H_2=\{(1),(1,2) \}$$ $$H_1\cap H_2=\{(1) \} $$ $$H_1\cup H_2=\{(1),(1,2),(1,3)\} $$

The union is not group since it is not closed as $(1,2)(1,3)=(1,3,2)\notin H_1\cup H_2$

for proof of inersection is a subgroup,just write; if you can take an element from a set you can do it,and follow the the other answers.

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  • $\begingroup$ thank you so much for you help :D $\endgroup$ – user131057 May 8 '14 at 0:20
  • $\begingroup$ @user131057: you are welcome. $\endgroup$ – mesel May 8 '14 at 7:39
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You can see the first from the definition of subgroup or the subgroup criterion quite easily. It's quite easy to make $H_1\cup H_2$ not a group... Consider two subgroups that intersect trivially. What happens when you add non-identity elements from these two subgroups?

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  • $\begingroup$ thank you .....i will consider your question:D $\endgroup$ – user131057 May 8 '14 at 0:21
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Perhaps a graphical representation will help you figure out the answer easier (it certainly did for me).

First of all, observe that if $H_1$ and $H_2$ are subgroups of $G$, then their intersection cannot be empty since it contains the identity element. Thus $H_1\cap H_2$ contains the identity element.

enter image description here Now, if $h_1$ and $h_2$ lie in their intersection,

  1. $h_1h_2$ should lie in $H_1$ as both of them lie in $H_1$ and $H_1$ is closed under the group operation.

  2. $h_1h_2$ should lie in $H_2$ as both of them lie in $H_2$ and $H_2$ is closed under the group operation.

Therefore from 1 and 2, $h_1h_2$ also lies in $H_1\cap H_2$, making it closed under the group operation.

A similar argument follows for inverses and associativity, thus making $H_1\cap H_2$ satisfy the group axioms and a group.

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