Let G be an abelian group and $a,b\in G$ be two distinct elements with a and b or order $2$.

  1. Show that $H=\{e,a,b,ab\}$ forms a subgroup and write out its multiplication table.

  2. Justify why all the elements in $H$ must be distinct.

  3. Prove that $$K=\{g \in G : |g| \leq 2\}.$$ Show that this can fail if $G$ is not abelian.

I can do part 1 (show $H$ has associativity, inverses, closure, and the identity element), but for parts 2 and 3 how do I show distinctness, assume two are equal and make a contradiction? Because I can't see how that will get me somewhere. And for 3, where do I begin?

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    $\begingroup$ What is $K$ here? $\endgroup$ – Travis Willse Sep 20 '14 at 14:58
  • $\begingroup$ For (2), assuming two are equal and deriving a contradiction would be efficient---there are only four elements, so there are very few cases, and all of them are easy. $\endgroup$ – Travis Willse Sep 20 '14 at 14:59
  • $\begingroup$ K is some group that has been created. Dont know why $\endgroup$ – Jack Armstrong Sep 20 '14 at 15:00
  • $\begingroup$ also are those cases for 2, e=a, a=b, and a=ab? $\endgroup$ – Jack Armstrong Sep 20 '14 at 15:00
  • $\begingroup$ Along with $ab = e$, and the observation that the $e = b$ and $b = ab$ cases follow by symmetry. $\endgroup$ – Travis Willse Sep 20 '14 at 17:32

$G$ have $4$ elements, one of which must be the unit $e$ and there must be another element $e\neq a\in G$.

Now, there is another element in the group $b$ with $b\neq a$ and $b\neq e$ .

Finally, consider the element $ab\in G$: $$ ab=a\implies a^{-1}ab=a^{-1}a\implies b=e $$

which is false. $$ ab=b\implies abb^{-1}=bb^{-1}\implies a=e $$

which is also false.

and $$ ab=e\implies b=a^{-1} $$

but $a$ is of order $2$ so $a=a^{-1}$ and thus $b=a$ which is again false.

We conclude $ab$ is the fourth element in $G$.

If I understood what part $3$ is asking you: Consider the Dihedral group of a square $G=D_{8}$,

We have it that $D_{8}=\langle s,rs\rangle$ since $r=rs\cdot s\in\langle s,rs\rangle$ but although $s,rs$ are both of order $2$ $$ G\neq K=\{g\in G\mid|g|\leq2\} $$

since $r\in D_{8}$ is of order $4$.

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  • $\begingroup$ @Jack - it is uncleared what asked, but I added an answer for that assuming I guessed what the question meant $\endgroup$ – Belgi Sep 20 '14 at 15:13
  • $\begingroup$ i saw ythat right after i posted that comment. but could you just explain it in more words? $\endgroup$ – Jack Armstrong Sep 20 '14 at 15:13
  • $\begingroup$ @Jack -What don't you understand ? $\endgroup$ – Belgi Sep 20 '14 at 15:14
  • $\begingroup$ after: the but althought s,rs are both of order 2, G is not equal to K $\endgroup$ – Jack Armstrong Sep 20 '14 at 15:17
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    $\begingroup$ @Jack - If $G$ is abelian and $a,b$ are distinct of order $2$ then then $H=\langle a,b\rangle$ is a subgroup of order $4$. But if $G$ is not abelian then $\langle a,b\rangle$ can have more elements in it, as I demonstrated using $D_{8}$ . If that not what part 3 wanted you to do, then I don't understand the question itself $\endgroup$ – Belgi Sep 20 '14 at 15:24

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