Find groups that contain elements $a$ and $b$ such that $|a|=|b|= 2$ and $|ab|=5$

My thoughts:
$|a|=|b|=2\implies a^2=e$ and $b^2=e$
I see that the group cannot be abelian as the order wont be greater than $2$ : $(ab)^2 = a^2b^2=e$
Not really sure how to proceed further. Greatly appreciate any help. Thanks!

  • $\begingroup$ Does "Find groups" mean "Find all groups" or "Find some groups"? Are you looking for a characterization or an example? $\endgroup$ – bof Aug 29 '15 at 23:14

Hint: A regular pentagon can be rotated and reflected

  • $\begingroup$ Thank you, $D_5$ has $10$ elements with $5$ rotations and $5$ reflections. All reflections have order $2$ and all rotations have order $5$... I am still not seeing the connection sorry.. should I take $a=b=$ reflection ? $\endgroup$ – pooja Aug 29 '15 at 8:33
  • $\begingroup$ @pooja Let $a$ be any reflection, let $c$ be any rotation. Find $b$ such that $ab=c$. Is $b$ a reflection or a rotation? $\endgroup$ – Hagen von Eitzen Aug 29 '15 at 9:13
  • $\begingroup$ Oh then i think $b$ must be a reflection because reflection composed with rotation can never give a rotation. $\endgroup$ – pooja Aug 29 '15 at 9:17

Here's one example: in the symmetric group $S_5$ $$(1\ 2\ 3\ 4\ 5) = (1\ 2)(3\ 5)\cdot(2\ 5)(3\ 4),$$ so take $a=(1\ 2)(3\ 5)$ and $b=(2\ 5)(3\ 4).$

You can get many more examples like this using the fact that any permutation can be expressed as the product of two involutions. So, for another concrete example, $$(1\ 2\ 3\ 4\ 5)(6\ 7\ 8\ 9\ 10)=(1\ 2)(3\ 5)(6\ 7)(8\ 10)\cdot(2\ 5)(3\ 4)(7\ 10)(8\ 9).$$


Hints: Suppose the group is finite, what do you know about its order?

Try writing the relations in terms of $a$ and $c=ab$. I use $1$ for the identity below.

You should get $a^2=c^5=1$. The third relation $b^2=1$ can be rewritten by noting that $b=a^2b=ac$ so that $acac=1$ and $ca=a^2cac^5=a(acac)c^4=ac^4$ so that every element can be written in the form $a^pc^q$ with $0\le p\lt 2$ and $0\le q \lt 5$

  • $\begingroup$ order of a group = number of elements in it ? with $a, ab$ as generators i think just get a group with 3 elements : $$\{e, a, ab\}$$ pardon for my ignorance.. i have just started algebra.. $\endgroup$ – pooja Aug 29 '15 at 8:38
  • 1
    $\begingroup$ @pooja Not at all - you have $c^5=1$ so the elements $c, c^2, c^3, c^4$ are in the group. Can you see how to use the relation $b^2=1$ to get something in terms of $a$ and $c$ - best try yourself first before looking at the answer. $\endgroup$ – Mark Bennet Aug 29 '15 at 8:56

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