# Let G be a group. Show that Gneq HcupKG =HcupK for any two subgroups Hleq GHleq G and Kleq GKleq G.

necessaryh 2020-11-06 Answered
Let G be a group. Show that $G\ne qH\cup KG=H\cup K$ for any two subgroups $H\le GH\le G\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}K\le GK\le G$.
You can still ask an expert for help

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it

## Expert Answer

delilnaT
Answered 2020-11-07 Author has 94 answers
Let G is a group and H,K are two non trivial subgroup of G. If possible, let $G=H\cup K$. Now Since, H not equal to G and K not equal to G, so we must have HH not equal to K and H and K can not be subset of each other. Therefore there exists $h\in H\setminus K\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}k\in K\setminus H$. Now since G is a group and $h,k\in H\cup K=G$, then $hk\in G=H\cup K$. So, hk is either in H or in K. Suppose $hk\in H$, since H is a subgroup and h∈H, so $h-1\in H.{h}^{-1}\in H$. Therefore,
${h}^{-1}\left(hk\right)=\left({h}^{-1}×h\right)k=k\in H$, which is against our choice, contradiction. Similarly, we can show that hk can not be in K. Henc, $G$ not equal to $H\cup K$, for non trivial subgroups H and K.
###### Not exactly what you’re looking for?

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it