Difference between revisions of "Generated subgroup"
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(Created page with "A cyclic subgroup is a group generated by a single element. ==Definition== Let {{M|(G,\times)}} be a group, and {{M|\{g_1,\cdots,g_n\}\subset G}} be a set...") |
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Claim: {{M|1=\langle\{g_i\}_{i=1}^n\rangle}} is a subgroup of {{M|G}} | Claim: {{M|1=\langle\{g_i\}_{i=1}^n\rangle}} is a subgroup of {{M|G}} | ||
{{Begin Proof}} | {{Begin Proof}} | ||
| + | To prove that {{M|1=\langle\{g_i\}_{i=1}^n\rangle}} is a subgroup we must show: | ||
| + | # The identity is in {{M|1=\langle\{g_i\}_{i=1}^n\rangle}} | ||
| + | # {{M|1=\langle\{g_i\}_{i=1}^n\rangle}} is closed (that is for {{M|1=a,b\in\langle\{g_i\}_{i=1}^n\rangle}} we have {{M|1=ab\in\langle\{g_i\}_{i=1}^n\rangle}}) | ||
| + | # {{M|1=\forall x\in\langle\{g_i\}_{i=1}^n\rangle[x^{-1}\in\langle\{g_i\}_{i=1}^n\rangle]}} | ||
| + | |||
| + | '''1) Proof that the identity is in {{M|1=\langle\{g_i\}_{i=1}^n\rangle}}''' | ||
| + | : This is trivial, as {{M|1=k=0}} in the definition "generates" the identity from {{M|1=\{g_i\}_{i=1}^n}} | ||
| + | |||
| + | |||
| + | '''2) Proof that {{M|1=\langle\{g_i\}_{i=1}^n\rangle}} is closed''' | ||
| + | {{Todo|Proof}} | ||
| + | |||
| + | |||
| + | '''3) Proof that for each {{M|1=x\in\langle\{g_i\}_{i=1}^n\rangle}} we have {{M|1=x^{-1}\in\langle\{g_i\}_{i=1}^n\rangle}} | ||
| + | {{Todo|Proof}} | ||
{{End Proof}}{{End Theorem}} | {{End Proof}}{{End Theorem}} | ||
{{Begin Theorem}} | {{Begin Theorem}} | ||
Latest revision as of 12:39, 12 May 2015
A cyclic subgroup is a group generated by a single element.
Contents
[hide]Definition
Let (G,×) be a group, and {g1,⋯,gn}⊂G be a set of elements of G, then the subgroup generated by {gi}ni=1[1] is given by:
- ⟨g1,⋯,gn⟩={hp11hp22⋯hpkk|k∈N0, hi∈{gj}nj=1, pi∈{−1,1}}
- Where it is understood that for k=0 the result of the operation on the empty list is e - the identity element of G
Informally that is to say that ⟨{gi}ni=1⟩ is the group that contains all compositions of the gi and their inverses, until it becomes closed under composition. This can be done because the gi∈G so 'worst case' if you will is that they generate a subgroup equal to the entire group
Proof of claims
[Expand]
Claim: ⟨{gi}ni=1⟩ is a subgroup of G
[Expand]
Claim: ⟨{gi}ni=1⟩ is a normal subgroup of G
TODO: Prove claims
