# Proof regarding the union of two subgroups (abstract algebra)

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?

• 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. – user50948 Oct 17 '15 at 16:26

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 $h_1h_2\in H_2$
• @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 $h_1^{-1}(h_1h_2)=h_2 \in H_1$. – Anurag A Oct 17 '15 at 17:38