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Given a group $G$ and two subgroups $H_1\leq G$ and $H_2\leq G$.

Also: $H_1\cup H_2= G$.

I have to prove, that either $H_1=G$ or $H_2=G$.

So, if the group $G$ is the union of the two subgroups $H_1$ and $H_2$, I must prove, that either $H_1$ or $H_2$ are trivial groups, am I correct?

But how would I do that?

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  • $\begingroup$ Your reformulation is not correct. There can definitely be two subgroups whose union is all of G, and yet neither of them is trivial. For instance take H_1 = G = H_2. $\endgroup$ – user50948 Oct 17 '15 at 16:26
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Claim: if $H_1\cup H_2$ is a group, then either $H_1\subseteq H_2$ or $H_2\subseteq H_1$.

If neither $H_1\subseteq H_2$ nor $H_2\subseteq H_1$ is true, then there is an $h_1\in H_1$ which is not in $H_2$, and a $h_2\in H_2$ which is not in $H_1$.

If $H_1\cup H_2$ is a group, then the product $h_1h_2$ is in $H_1\cup H_2$. Thus the product is in $H_1$ or in $H_2$. If $h_1h_2\in H_1$, then $h_2\in H_1$, contradicting the choice of $h_2$. We get a similar contradiction if $h_1h_2\in H_2$

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  • $\begingroup$ I don't understand your last two sentences. If h_1 h_2 are in H_1, why is h_2 in H_1? Before that, you have declared that h_2 is in H_2 $\endgroup$ – de_dust Oct 17 '15 at 17:22
  • $\begingroup$ @de_dust recall that$H_1$ is a group so if $h_1 \in H_1$, then $h_1^{-1} \in H_1$. So now if $h_1h_2 \in H_1$, then $h_1^{-1}(h_1h_2)=h_2 \in H_1$. $\endgroup$ – Anurag A Oct 17 '15 at 17:38
  • $\begingroup$ @AnuragA: Now I understand. Thanks to both of you! $\endgroup$ – de_dust Oct 17 '15 at 20:12

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