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If there is a group $G$ with order $a$, having a subgroup $H_1$ with order $b$, and $H_2$ with order $c$, and $bc=a$, $H_1 \cap H_2 = e $. Is $H_1 H_2 =G$?

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    $\begingroup$ If $G$ is finite, so is $H_1$ and $H_2$. We have a formula $$\frac{|H_1| \cdot |H_2|}{|H_1 \cap H_2|} = |H_1 H_2| \ . $$ Can you continue? $\endgroup$ – Gustavo Mar 30 '16 at 5:06
  • $\begingroup$ See math.stackexchange.com/a/903199/210479 $\endgroup$ – sqtrat Mar 30 '16 at 5:06
  • $\begingroup$ thanks for the idea. Its like an answer below give a complete explanation. $\endgroup$ – Agus Ahmad Rizqi Mar 30 '16 at 9:29
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The answer to your question is yes: if $G$ is a finite group, and $H$ and $K$ are subgroups such that $|G| = |H||K|$ and $H \cap K = 1$, then $G = HK$.

The easiest way to see this is to use the identity $$|HK| = \frac{|H||K|}{|H \cap K|}$$ Note that this identity holds even if $HK$ is merely a subset (not a subgroup) of $G$.

In your case, $|H \cap K| = 1$ and $|H||K| = |G|$, so the identity becomes $$|HK| = |G|$$ Therefore, $HK$ is a subset of $G$ with the same cardinality as $G$, so $HK = G$.


For completeness, here is a simple proof of the identity.

Recall that $H \times K$ is the direct product of $H$ and $K$. It is the set of all ordered pairs of the form $(h,k)$ where $h \in H$ and $k \in K$, and it has cardinality $|H \times K| = |H||K|$. ($H \times K$ inherits a group operation from $H$ and $K$, but we won't need to use it here.)

Also recall that $HK$ is the set $\{hk : h \in H, k \in K\}$ which is not necessarily a subgroup of $G$, but it is certainly a subset of $G$.

Define the map $f : H \times K \to HK$ by $f(h,k) = hk$. Clearly this map is surjective, so its image is $HK$, which has cardinality $|HK|$.

Now, how many elements of $H \times K$ are mapped to a given $hk \in HK$?

Note that if $d \in H \cap K$, then $(hd, d^{-1}k) \in H \times K$, and $f(hd, d^{-1}k) = hk$. This shows that there are at least $|H \cap K|$ elements of $H \times K$ which are mapped to $hk$.

It's easy to verify that these are all of the elements which are mapped to $hk$. Suppose that $f(h', k') = h'k' = hk$. Then $h^{-1}h' = k(k')^{-1} \in H \cap K$, call this element $d$. Thus $h' = hd$ and $k' = d^{-1}k$.

This shows that for each $hk \in HK$, there are exactly $|H \cap K|$ elements of $H \times K$ which are mapped to $hk$. Consequently, $$|H \times K| = |H \cap K||HK|$$ Since $|H \times K| = |H||K|$, this is equivalent to $$|H||K| = |H \cap K| |HK|$$ Dividing by $|H \cap K|$ gives us the desired identity.

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  • $\begingroup$ Thanks for the idea. But, i still confuse about why $H_1 H_2$ should be a subgroup? $\endgroup$ – Agus Ahmad Rizqi Mar 30 '16 at 9:30
  • $\begingroup$ @agus In general, it is not a subgroup, but in this case it turns out to be all of $G$, so it is a subgroup. $\endgroup$ – Bungo Mar 30 '16 at 15:34
  • $\begingroup$ I'm sorry, i'm a beginner in this course. What about a subgroup of finite group require $ab$ also in it? In $H_1 H_2$, if $h_1$ element of $H_1$, and $h_2$ element of $H_2$, then $H_1 H_2={h_1 h_2}$, and if it operated with other element $h_1^'h_2^'$, it yields $h_1 h_2 {h_1}^'{h_2}^'$. It seems not element of $H_1 H_2$. $\endgroup$ – Agus Ahmad Rizqi Mar 31 '16 at 5:23
  • $\begingroup$ @AgusAhmadRizqi: At the start of the problem we do not assume that $H_1 H_2$ is a subgroup. Without making this assumption, we are able to prove that $G = H_1 H_2$. Now $G$ is of course a group, so this means that $H_1 H_2$ is a group after all. $\endgroup$ – Bungo Mar 31 '16 at 5:26
  • $\begingroup$ oh, i understand. every element of $h_1 h_2$ should be in G. Thank you very much. It will be very useful for my paper next week. I'm an undergraduate student of mathematics program from indonesia, sorry if my english is not good. Again, thank you. $\endgroup$ – Agus Ahmad Rizqi Mar 31 '16 at 6:17
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Since $H_1$ and $H_2$ are subgroups of $G$, we have $$|H_1H_2|=\frac{|H_1||H_2|}{|H_1\cap H_2|}=|H_1||H_2|=|G|,\tag1 $$ so $H_1H_2=G$.

(See @Bungo's answer for a proof of the identity in (1))

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  • $\begingroup$ @Bungo Good catch, the counterexample was off-target. Let me revise. $\endgroup$ – Math1000 Mar 30 '16 at 5:20
  • $\begingroup$ In your counterexample, $H_1$ is not a subgroup, because it does not contain $\tau \sigma = (123)$. Indeed, the formula $|HK| = |H||K|/|H \cap K|$ shows that there is no counterexample if $H \cap K = 1$ and $|H||K| = |G|$, for then we have $|HK| = |G|$ and hence $HK = G$. $\endgroup$ – Bungo Mar 30 '16 at 5:21
  • $\begingroup$ P.S. The formula $|HK| = |H||K|/|H \cap K|$ is valid for any finite subgroups, whether abelian or not, and whether or not $HK$ is a subgroup. So the bit at the beginning where you first consider the abelian case seems unnecessary. $\endgroup$ – Bungo Mar 30 '16 at 5:28
  • $\begingroup$ @Bungo Indeed, I am not very well-versed in algebra ;) $\endgroup$ – Math1000 Mar 30 '16 at 5:31
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    $\begingroup$ No worries, I just want to avoid confusing the OP! $\endgroup$ – Bungo Mar 30 '16 at 6:45

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