E2 - lagrange's theorem

|H| divides |G|

let $a_{1} H, a_{2} H, \dots, a_{r} H$ denote the distinct left cosets of $H$ in $G$ , then $\exists a H = a_{i} H \forall a \in G$
also, $a \in a H$ , so, each member of $G$ belongs to one of the cosets $a_{i} H$ , ie:

G = a_{1} H \cup \dots \cup a_{r} H

| G | = | a_{1} H | + \dots + | a_{r} H |

finally, $| a_{1} H | = | H | \forall i ⟹ | G | = r | H |$
this theorem provides a list of candidates for the subgroups of a group, ie. subgroup candidate criterion
eg: if $| G | = 12$ , then it may have subgroups of order $12, 6, 4, 3, 2, 1$ only
note: the converse is false, ie. a $| G | = 12$ need not have a subgroup of order 6
eg: $A_{4}$ of order 12 has no subgroups of order 6

index

the number of distinct left cosets of $H$ in $G$ for $H \leq G$ , denoted by $| G : H |$

| G : H | = | G | / | H |

| a |

divides

| G |

in a finite group, the order of each element of the group divides the order of the group

groups of prime order are cyclic

suppose $G$ has a prime order, and $a \in G$ such that $a \neq e$ , then $| ⟨ a ⟩ |$ divides $P$ and $| ⟨ a ⟩ | \neq 1$ , so, $| ⟨ a ⟩ | = | G |$

a^{| G |} = e

$| G | = | a | k$ for some $k \in Z > 0$ , therefore, $a^{| G |} = a^{| a | k} = e^{k} = e$

fermat's little theorem

a^{p} mod p = a mod p

division algorithm: $a = p m + r$ , $0 \leq r < p ⟹ a mod p = r \equiv a$
hence, it suffices to prove that $r^{p} mod p = r \equiv a$
it is trivial for $r = 0$
$r \in U (p)$ as $p$ is a prime, and $| U (p) | = p - 1$
$r^{p - 1} mod p = 1$ , therefore $r^{p} mod p = r$
lagrange's theorem shows how the finiteness of a group imposes severe restrictions on the possible orders of subgroups
the next theorem is a counting technique that also puts limits on the existence of certain subgroups in finite groups

| H K | = | H | | K | / | H \cap K |

for two finite subgroups $H$ and $K$ of a group, define the set $H K = {h k ∣ h \in H, k \in K}$ , then, $| H K | = | H | | K | / | H \cap K |$

note: $H K$ may not be a subgroup
eg: $| G | = 75$ , and it can have at most one subgroup of order 25
suppose $H, K \leq G$ such that their orders are 25
since $| H \cap K |$ divides $| H | = 25$ , $| H \cap K | = 1$ or $5 ⟹ | H K | = | H | | K | / | H \cap K | = 625$ or $125$
thus, $| H \cap K | = 25$ , and therefore, $H = K$

classification of groups of order 2p

if $G$ is a group with $| G | = 2 p$ , where $p$ is a prime greater than $2$ , then $G$ is isomorphic to $Z_{2 p}$ or $D_{p}$