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If a group is generated by an element does that mean the generator commutes with all the other elements or does it mean that because the group is cyclic(as it has a generator) that all elements commute with each other.

For example, I am trying to find the conjugacy classes of the group D4 and am not sure if I could use the property that elements commute with each other. It seems to be taking too long so I was wondering what would be some facts I could be using?

From the notation below I understand that D4 is generated by a and b. So, is it only these two elements that commute with the others? $$ D_4=\langle a, b\rangle=\{e, a, a^2, a^3, b, ab, a^2b, a^3b\} $$

user1729
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chloe loughan
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  • To learn about groups you should work lots of examples, not just try to prove things. Wrtie out the multiplication tables for $D_3$ and $D_4$ to start with. This question may help: Thttps://math.stackexchange.com/questions/2205865/some-subgroup-of-dihedral-group-is-normal/2766448#2766448 – Ethan Bolker May 17 '18 at 10:02
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    While the main question has already been answered, it seems worthwhile to point out that the following properties are equivalent: "all elements commute (i.e. the group is abelian)", "all elements commute with all generators" and "all the generators commute with each other". – Tobias Kildetoft May 17 '18 at 10:52

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The short answer is: the idea of "generators" and "commutativity" are completely disjoint. I don't really know what else to say...

In $D_4$, the generators $a$ and $b$ do not commutate with each other, so cannot each commute with every element. For example, $ba=a^3b\neq ab$.

user1729
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I'm not sure if this answers your question, but here it goes.

If a group is generated by a single element $a$, then $a$ commutes with all other elements. In fact, the group is Abelian then.

But if a group is generated by more than one element, there is no reason to assume that the generators commute with all other elements; this happens if and only if the group is Abelian.

In particular, in $D_4$ it is not true that $a$ and $b$ commute with all other elements.

José Carlos Santos
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  • This has cleared up the confusion thanks. Is there a recommended way to find the conjugacy classes of a group in a rigorous approach? – chloe loughan May 17 '18 at 10:02