# The automorphism of a group when given a group presentation

How do you find the automorphism group when given a group presentation?

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At the level of generality at which you posed the question, you probably can't. Do you have a specific example in mind? –  Mariano Suárez-Alvarez Apr 30 '13 at 13:44
I do have a particular thing in mind, but it still is quite general: $<x_{1},x_{2} , \ldots , x_{n} | x_{i} = w_{i} x_{p(i)} w_{i}^{-1} , i = 1 , 2 , \ldots , n >$ where $p(1) , p(2) , \ldots , p(n)$ are a permutation of $1 , 2 , \ldots , n$ and $w_{i} (i = 1 , 2 , \ldots , n)$ are words in $x_{1} , x_{2} , \ldots , x_{n}$ which satisfy the identity $\prod_{i = 1}^{n} x_{i} = \prod_{i = 1}^{n } w_{i} x_{p(i)} w_{i}^{-1}$ in the free group $<x_{1} , x_{2} , \ldots , x_{n} >$ –  Jesse Apr 30 '13 at 13:51
Given that the automorphism group is trivial iff the group is, the general problem is not decidable. –  Tobias Kildetoft Apr 30 '13 at 13:51
@TobiasKildetoft ...or $C_2$... –  user1729 Apr 30 '13 at 13:51
@user1729 Thank you. Not sure how I forgot that. I think it still makes it undecidable. –  Tobias Kildetoft Apr 30 '13 at 13:52