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Though often accepted as a basic fact, can it be proved that $\{e, g, g^2,\ldots ,g^k\}$ is a subgroup of $G$, where $(G, *)$ is a finite group, $g$ is an element of $G$, and $k$ is the order of $g$? I see this as definitions everywhere, but I cannot find any proof that this is in fact, true.

To clarify, $"*"$ represents an generic binary operation, not multiplication, and $g^2 = g * g.$

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    What are the conditions to be a subgroup? What have you tried showing so far? – 2017-02-20
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    You should verify that the subset you defined is closed under $*$, and that each of its elements has an inverse in the same subset. Did you try to show these? – 2017-02-20
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    You need to go through the steps of a subgroup test . Do you know how to do that ? – 2017-02-20
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    well, I know that there exists a positive k such that g^k = e by definition of the order of an element. – 2017-02-20
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    Note that finiteness is incredibly important here. This is not necessarily true for an infinite group. – 2017-02-20
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    https://en.wikipedia.org/wiki/Subgroup_test – 2017-02-20

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First of all: if the order of $g$ is $k$, then we have that $g^k = e$, so that your group actually is $H = \{e, g, g^2, \ldots, g^{k-1}\}$. Are you aware of the subgroup criterion? (If no, it states that if $H$ is a subset of a group $G$ and $H$ is not empty, then it suffices to prove that for any $x, y \in H$ we also have $x \ast y^{-1} \in H$ in order to prove that $H$ is a subgroup. You can easily prove this yourself).

Now clearly $H$ is not empty, since $e \in H$. Now suppose $x^l, x^m \in G$, where both $l,m \in \{0, 1, \ldots k-1\}$. Consider $x^l \ast (x^m)^{-1}$. Note that $(x^m)^{-1} = x^{-m}$ (you can prove this, but I guess this is clear).

Now we distinguish two cases: either $l \geq m$ or $l < m$. In the first case we have that $x^l \ast x^{-m} = x^{l - m} \in G$ since $0 \leq l - m \leq k-1$.

If $l < m$, then we would have that $x^{l-m}$ has a negative exponent. However, we have that $x^k = e$, so we find that $x^l \ast x^{-m} = x^{l-m} = x^{l-m} \ast x^k = x^{l-m + k}$ and since $l

By the subgroup criterion, we have that $H = \{e, g, g^2, \ldots, g^{k-1}\}$ is a subgroup.

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    I think this depends on the instructor you have: I learned about the subgroup criterion as the first theorem after the definition of a subgroup and the statement that the neutral element of the subgroup corresponds to the neutral element of the group (same for inverse elements). However, I also think that I was wrong in assuming that the OP should have seen, so thank you to point this out to me. This will make me a better explainer :) – 2017-02-20
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Let $H:= \{g^n \mid g \in G, n \in \mathbb N\}$.

Let $g \in G$. If $G$ is finite, then there exists some $k \in \mathbb N$ so that $g^k=e$, since otherwise $|G|= \infty$. So, identity is in the subgroup, and it is also enough to consider $$H=\{g^n \mid 1 \leq n \leq k\}$$

For each $1 \leq n \leq k$, we can take $g^n \cdot g^{k-n}=g^k=e$, but we know that $g^{k-n} \in H$, so that inverses exist.

You should show that the group is closed under multiplication.

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    Very nicely written; concise, and also thorough (and leaving a little work for the OP). – 2017-02-20
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    I edited the question to clarify. * does not denote multiplication, but rather an operation on these elements – 2017-02-20
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    @amWhy thank you! I admire your writing a lot, so I appreciate it. – 2017-02-20
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    @ReiHinoX $\cdot$ is just convention in group theory, one could have just as easily written $+$ and asked a question about $ng$ instead of $g^n$. The multiplication is taken to be generic. – 2017-02-20