combinatorics: improved the FpGroups.index() method

[Ford2003]

combinatorics: FpGroup.index() optimisiation

References to other Issues or PRs

Brief description of what is fixed or changed

Added the use of Lagrange's theorem to FpGroup.index() method so that it returns |G|/|H| when G has finite order, by using FpGroup.subgroup() and FpGroup.order() methods.

Other comments

I think there is an issue with this method, if the FpGroup has infinite order, but is not picked up by FpGroup._is_infinite() method. I can create this issue with an example.
Added myself to .mailmap as I'm new contributor.

Release Notes

• combinatorics
  • Added a small optimisation to FpGroup.index() method to return |G|/|H| if G has finite order.

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• combinatorics
  • Added a small optimisation to FpGroup.index() method to return |G|/|H| if G has finite order. (#26458 by @Ford2003)

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combinatorics: FpGroup.index() optimisiation

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#### Brief description of what is fixed or changed
Added the use of Lagrange's theorem to FpGroup.index() method so that it returns |G|/|H| when G has finite order, by using FpGroup.subgroup() and FpGroup.order() methods. 

#### Other comments
I think there is an issue with this method, if the FpGroup has infinite order, but is not picked up by FpGroup._is_infinite() method. I can create this issue with an example.
Added myself to .mailmap as I'm new contributor.

#### Release Notes

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* combinatorics
  * Added a small optimisation to FpGroup.index() method to return |G|/|H| if G has finite order.
<!-- END RELEASE NOTES -->

[sylee957]

Is order always terminating? I'm aware that some fpgroups algorithms may not terminate, or order can be fairly slow, so I'm not sure about this

[Ford2003]

I think it doesn't always because an FpGroup can have infinite order, but isn't picked up by _is_infinite method. So perhaps this is something that is added once the _is_infinite method is improved?