Difference between revisions of "Generated subgroup"
(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.
Definition
Let [ilmath](G,\times)[/ilmath] be a group, and [ilmath]\{g_1,\cdots,g_n\}\subset G[/ilmath] be a set of elements of [ilmath]G[/ilmath], then the subgroup generated by [ilmath]\{g_i\}_{i=1}^n[/ilmath][1] is given by:
- [math]\langle g_1,\cdots,g_n\rangle=\{h_1^{p_1}h_2^{p_2}\cdots h_k^{p_k}|k\in\mathbb{N}_0,\ h_i\in \{g_j\}_{j=1}^n,\ p_i\in\{-1,1\}\}[/math]
- Where it is understood that for [ilmath]k=0[/ilmath] the result of the operation on the empty list is [ilmath]e[/ilmath] - the identity element of [ilmath]G[/ilmath]
Informally that is to say that [math]\langle\{g_i\}_{i=1}^n\rangle[/math] is the group that contains all compositions of the [ilmath]g_i[/ilmath] and their inverses, until it becomes closed under composition. This can be done because the [ilmath]g_i\in G[/ilmath] so 'worst case' if you will is that they generate a subgroup equal to the entire group
Proof of claims
Claim: [ilmath]\langle\{g_i\}_{i=1}^n\rangle[/ilmath] is a subgroup of [ilmath]G[/ilmath]
To prove that [ilmath]\langle\{g_i\}_{i=1}^n\rangle[/ilmath] is a subgroup we must show:
- The identity is in [ilmath]\langle\{g_i\}_{i=1}^n\rangle[/ilmath]
- [ilmath]\langle\{g_i\}_{i=1}^n\rangle[/ilmath] is closed (that is for [ilmath]a,b\in\langle\{g_i\}_{i=1}^n\rangle[/ilmath] we have [ilmath]ab\in\langle\{g_i\}_{i=1}^n\rangle[/ilmath])
- [ilmath]\forall x\in\langle\{g_i\}_{i=1}^n\rangle[x^{-1}\in\langle\{g_i\}_{i=1}^n\rangle][/ilmath]
1) Proof that the identity is in [ilmath]\langle\{g_i\}_{i=1}^n\rangle[/ilmath]
- This is trivial, as [ilmath]k=0[/ilmath] in the definition "generates" the identity from [ilmath]\{g_i\}_{i=1}^n[/ilmath]
2) Proof that [ilmath]\langle\{g_i\}_{i=1}^n\rangle[/ilmath] is closed
TODO: Proof
3) Proof that for each [ilmath]x\in\langle\{g_i\}_{i=1}^n\rangle[/ilmath] we have [ilmath]x^{-1}\in\langle\{g_i\}_{i=1}^n\rangle[/ilmath]
TODO: Proof
Claim: [ilmath]\langle\{g_i\}_{i=1}^n\rangle[/ilmath] is a normal subgroup of [ilmath]G[/ilmath]
TODO: Prove claims
