Cardinality of a group Let $G = <a,b,c>$, where $a=(1,2)$, $b=(2,3)$ and $c=(4,5)$ are elements of $S_5$
Basically I had a question to compute the orbit and stabilisers and eventually state the orbit-stabiliser theorem and by that theorem I found the Cardinality of |G| to be 12. 
My question may be trivial but how can I find the cardinality of G without doing that?(as if to confirm I'm correct)

Well, we have that
$$a^2=b^2=c^2=1\;,\;\;(ab)^3=1\;,\;\;ac=ca\;,\;\;bc=cb$$
Thus, we already have that $\;G\;$ cannot be abelian ($\;ab\neq ba\;$) and it has a center of at least order $\;2\;$ .
Also, check that any appearance of the generator $\;c\;$ in some word (element) of the group can be "dragged" to the beginning (or end) of the word, and all other letters in the word (only $\;a,b\;$) are either of the form $\;ab...\;,\;\;ba...\,\;$, and with no more than two consecutive $\;abab\;,$ or $\;baba\;$ (as $\;ab\;,\;\;ba\;$ are conjugate and thus have the same order...
From here, with some basic albeit somehow boring combinatorics, you can reach the number of elements in $\;G\;$ without any group action, stabilizers and stuff.
