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Say S is x and y. x is an element of order 2. y is an element or order 3. So

$x^2=e$ $y^3=e$

say x and y is unrelated. So x is not a member of subgroup generated by y and via versa.

How many elements will $<S>$ have?

$e,x,y,y^2,xy, xy^2,...$ and so on

Is there a limit?

Must $<S>$ be finite?

What's the limit?

Is there a general rule on how to compute order of group generated by these 2 simple elements.

Also do all groups generated by 2 elements, with 2 and 3 cycle each, isomorphic to this group?

user4951
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    Your first line is poorly worded and not understandable. Your last sentence is not a sentence. Please formulate and format questions carefully before expecting others to spend their time deciphering it. – P Vanchinathan Nov 10 '16 at 08:14
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    If I guessed the questions correctly, the answers are: No. No. Infinite. No. But, to get any kind of helpful details, you should give your own thoughs also. I realize that you have difficulties with English, but please try. – Jyrki Lahtonen Nov 10 '16 at 08:16
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    It can be infinite. An example is the group generated by $x\mapsto-1/x$ and $x\mapsto(x+1)/x$. See http://math.stackexchange.com/questions/927879/generating-the-special-linear-group-of-2-by-2-matrices-over-the-integers – Gerry Myerson Nov 10 '16 at 08:17
  • It can be infinite? And we don't know? I think the rule is pretty well defined there. All other groups that is also generated by 2 elements will be isomorphic to that group right. So there should be a definite answer, even though it's infinite. Is it infinite or not? – user4951 Nov 10 '16 at 08:20
  • Jim Thio, it depends on the group the elements $x$ and $y$ come from. If they are elements of $S_3$, or $C_6$, yes then the group is finite. But with no other information (i.e. relations that $x$ and $y$ may satisfy) it can be infinite. If you know that there are no other relations, then it is infinite. The infinite example I had in mind is isomorphic to Gerry's example. But, if you know for example that $xy=yx$, then the order is at most six. Same, if $xyx^{-1}=y^{-1}$. Without knowing whether such relations hold all bets are off. – Jyrki Lahtonen Nov 10 '16 at 08:25
  • The last question can be answered as follows: any group generated by such elements $x$ and $y$ is a quotient group of the free product $C_2*C_3$. And that free product has infinitely many elements. Some of its quotients are finite though. – Jyrki Lahtonen Nov 10 '16 at 08:26
  • what is quotient group? – user4951 Nov 10 '16 at 08:29
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    oh I see. Imagine 2 and 3 with rules of addition modulus 6. Then there are only 6 elements. However, there is no limit on how many members of the group in general case are. – user4951 Nov 10 '16 at 08:30
  • Two years to late. But I interpreted these to mean $S={x,y}$ where $x,y$ are members of some group $G$ where $|x|=2;|y|=3$. And $$ is the subgroup of $G$. The answer is very simply that $$ has 6 elements and yes, it must be be finite and it must have the lowest common multiple of elements. and yes all such groups are isomorphic this group. – fleablood Jun 09 '18 at 15:56
  • @fleablood: I don't understand your comment. Gerry and Jyrki and I have given examples contradicting it. – user21820 Jun 11 '18 at 07:48

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Hint: Consider the euclidean plane. Let $P,Q$ be any two distinct points. Let $H$ be a half-rotation about $P$. Let $T$ be a third-rotation about $Q$. Then prove that $H,T$ generate an infinite group of transformations of the plane. (Show that you can use $H,T$ to move $P$ arbitrarily far away.)

user21820
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