Generated subgroup

A cyclic subgroup is a group generated by a single element.

Definition

Let $(G,\times)$ be a group, and $\{g_1,\cdots,g_n\}\subset G$ be a set of elements of $G$, then the subgroup generated by $\{g_i\}_{i=1}^n$ is given by:

• $$\langle g_1,\cdots,g_n\rangle=\{h_1^{p_1}h_2^{p_2}\cdots h_k^{p_k}|k\in\mathbb{N}_0,\ h_i\in \{g_j\}_{j=1}^n,\ p_i\in\{-1,1\}\}$$

  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

$$\langle\{g_i\}_{i=1}^n\rangle$$ is the group that contains all compositions of the $g_i$ and their inverses, until it becomes closed under composition. This can be done because the $g_i\in G$ so 'worst case' if you will is that they generate a subgroup equal to the entire group

Proof of claims

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Claim: $\langle\{g_i\}_{i=1}^n\rangle$ is a subgroup of $G$

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To prove that $\langle\{g_i\}_{i=1}^n\rangle$ is a subgroup we must show:

1. The identity is in $\langle\{g_i\}_{i=1}^n\rangle$
2. $\langle\{g_i\}_{i=1}^n\rangle$ is closed (that is for $a,b\in\langle\{g_i\}_{i=1}^n\rangle$ we have $ab\in\langle\{g_i\}_{i=1}^n\rangle$)
3. $\forall x\in\langle\{g_i\}_{i=1}^n\rangle[x^{-1}\in\langle\{g_i\}_{i=1}^n\rangle]$

1) Proof that the identity is in $\langle\{g_i\}_{i=1}^n\rangle$

This is trivial, as $k=0$ in the definition "generates" the identity from $\{g_i\}_{i=1}^n$

2) Proof that $\langle\{g_i\}_{i=1}^n\rangle$ is closed

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TODO: Proof

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3) Proof that for each $x\in\langle\{g_i\}_{i=1}^n\rangle$ we have $x^{-1}\in\langle\{g_i\}_{i=1}^n\rangle$

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TODO: Proof

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Claim: $\langle\{g_i\}_{i=1}^n\rangle$ is a normal subgroup of $G$

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TODO: Prove claims

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See also

• Cyclic subgroup
• Commutator
• Conjugation
• Normal subgroup