# If there for a finite group exists a finite subgroup of less order does it have to have a coset?

If there for a finite group G exists a subgroup H with lower order than G does H has to have a coset?

My thinking is that it has to be the case, as the elements not in H at least for one subgroup, that group having to be a coset. Am I completly off?

Yes, every subgroup has cosets. Will come back to this later.

In the second paragraph you say " as the elements not in H at least for one subgroup, that group having to be a coset". It sounds like you think that some elements outside of $H$ can form a subgroup. This is never true. A subgroup must by definition contain the identity $e$, so any two subgroups have at least one element in common, $e$. Thus it's impossible for a subgroup to be totally from outside another subgroup.

A coset is a subset of the group of the form $gH$, where $g$ is any element in $G$. Thus $eH = H$ is a coset, so there is at least one coset. If you another element $g$ outside of $H$, $gH$ is a coset as well.

• But can you guarantee that all cosets in total cover the whole of G? – User123456789 Sep 18 '18 at 17:24
• Note that if $g$ times a member of $H$ is another member of $H$, then $gh_1=h_2$, therefore $g=h_2h_1^{-1}$. But if $H$ is a subgroup, then it is closed under inverses and multiplication, so $h_2h_1^{-1}$ is in $H$. Thus, if $g$ is not in $h$, $gH$ is distinct from $H$. – Acccumulation Sep 18 '18 at 17:30
• @User123456789 Yes. Given any element $g$ of $g$, $g$ is trivially in the coset $gH$. This is because for $H$ to be a subgroup, it must contain the identity, so $ge$ is in $gH$. Note that when listing the cosets, $gH$ might not explicitly appear written in that manner; there may be some $g_2$ such that $g_2H$ is listed and $g_2H$ is the same as $gH$. This implies that $gg_2^{-1}$ is in $H$. – Acccumulation Sep 18 '18 at 17:34
• And thus as $e$ is in $H$ one can guarantee that every element in $G$ must be in one and only one coset of $H$? Edit: See that you answered it in the comment above. Thanks! – User123456789 Sep 18 '18 at 17:34

A (left) coset of $H$ is precisely a set of the form $gH=\{gh|h\in H\}$ where $g\in G$, so $H=1_GH$ (where $1_G$ is the identity element of $G$) is itself a coset.

Note that there is no restriction on the order of $H$, so $G$ is also a coset (of $G$)