Exercise 3.23 a&c
Can we assume that G is a group? and futhermore can we assume G is a cyclic group?
Exercise 3.23 a&c
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» Exercise 3.23 a&c
Yes, you should assume that $G$ is a group. However, you should not assume that $G$ is cyclic (and you don't need this fact).
There are many paths one could take to prove this. Here's one approach.
Assume that $|a|=n$. You need to show that $|a^{-1}|=n$. Certainly, $a^{-1}$ has an order; suppose that it is $k$. This implies that $(a^{-1})^k=e$ and that this is the smallest natural number for which this happens. Does this imply anything about $a$? What relationship is there between $n$ and $k$? First, you should conclude that $k$ is a multiple of $n$ (by Proposition 3.5 applied to the subgroup generated by $a$). Next, compute $(a^{-1})^n$. What can you conclude about $k$? (Recall $k$ was the minimal natural number with some property.)
Alright thanks that really helps, and would you also find the smallest possible order of $a$ with an arbitrary element as the order. Then compare how $a$ and $a^{-1}$ have the same order?
Jess, I'm not sure what you're asking. Are you suggesting an alternate approach or asking if you need to do something in addition to what I asked?
Allow me to be picky about something you said since it might clear up some possible confusion. You said:
and would you also find the smallest possible order of $a$
The order of an element is already as small as possible, by definition. You continued:
with an arbitrary element as the order.
Group elements can be all sorts of things: numbers, matrices, symmetries, motions, functions, etc. However, the order of a group element is always a natural number.
So, we can't assume in 23c that the group is cyclic? If we can't, how can we begin to talk about commutativity? Wow, is that spelled correctly?
Cathy Fulkerson Ajamie
Correct, you cannot assume that $G$ is cyclic and you don't need this. However, the subgroup generated by $\langle a \rangle$ is cyclic and you can apply any theorems about cyclic groups to $\langle a \rangle$.
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