Product of Subgroup with Itself

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {G, \circ}$ be a group.


Then:

$\forall H \le G: H \circ H = H$

where $\le$ denotes the subgroup relation.


Proof

From Magma Subset Product with Self, we have:

$H \circ H \subseteq H$


Let $e$ be the identity of $G$.

By Identity of Subgroup, it is also the identity of $H$.


So:

\(\ds h\) \(\in\) \(\ds H\)
\(\ds \leadsto \ \ \) \(\ds e \circ h\) \(\in\) \(\ds H \circ H\)
\(\ds \leadsto \ \ \) \(\ds h\) \(\in\) \(\ds H \circ H\)
\(\ds \leadsto \ \ \) \(\ds H\) \(\subseteq\) \(\ds H \circ H\)

Hence the result from the definition of set equality.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Product_of_Subgroup_with_Itself&oldid=623036"