Show these are subgroups of a group G?

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In summary, Z(G) is the center of a group G, defined as the set of elements in G that commute with all other elements in G. It is a subgroup of G, as proven in the link provided.C(g) is the centralizer of an element g in G, defined as the set of elements in G that commute with g. It is also a subgroup of G, with a similar proof to that of Z(G). These are two separate questions because Z(G) and C(g) are not the same thing.
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Firepanda
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Z(G) = { x in G : xg=gx for all g in G } (center of a group G)

and

C(g) = { x in G : xg=gx } (centralizer of g in G)Z(G) is a subgroup by the proof here:

http://en.wikipedia.org/wiki/Center_(group_theory)#As_a_subgroupHow do I go about showing C(g) is a subgroup? The proofs look like they should be identical to me, but then why do I have 2 separate questions for it?

Thanks
 
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They are two different questions because Z(G) is not the same thing as C(g). Yes, the proofs are almost identical. So just go ahead and prove C(g) is a group.
 

FAQ: Show these are subgroups of a group G?

What is a subgroup?

A subgroup is a subset of a group that satisfies the group axioms and forms a group itself under the same operation.

How do you show that a subset is a subgroup of a group?

To show that a subset H is a subgroup of a group G, we need to prove that it satisfies the subgroup criteria:

  • H is closed under the group operation
  • H contains the identity element of G
  • H contains the inverse of each of its elements in G

Can a group have multiple subgroups?

Yes, a group can have multiple subgroups. In fact, every group has at least two subgroups - itself and the trivial subgroup containing only the identity element.

How do you determine the order of a subgroup?

The order of a subgroup is equal to the number of elements in the subgroup. To determine the order, count the number of elements in the subset that satisfies the subgroup criteria.

Is the intersection of two subgroups also a subgroup?

Yes, the intersection of two subgroups is also a subgroup. This is because the intersection will also satisfy the subgroup criteria and form a group under the same operation.

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