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Let $A$ be a non-empty set and consider $S(A)$ the set of all bijective functions from $A$ to itself, with composition as group operation. Fix $a_0 \in A$. Define $H = \{f \in S(A): f(a_0) = a_0\}$. Prove that $H$ is a subgroup of $S(A)$.

does this tell me $a_0$ is an identity or that $f(a_0)$ is an identity? I am assuming I need to use something to do with subgroup theorems but I am having trouble understanding how to use these.

Ok that makes more sense since $f(a_0)$ = $a_0$ and $g(a_0)$ = $a_0$ we can conclude that $f^{-1}$ $f(a_0)$ = $f^{-1}$ $a_0$ so we know $f^{-1}$ is in the set? So $f^{-1}$ $g(a_0) $= $a_0$ and the by the theorem we may conclude $H$ is a subgroup of G?

As a side note would love to know WHERE you guys find the element signs and how to put subscripts on things if you could link me to a list of stuff you can put into here with some instructions.

Someone has pointed me a direction a few times but i have never found something I understood. I am not a programer but if I can find a lost of things I am sure I can figure it out usually I just search until I find a post with what I want click edit and take the symbols I want but if theirs instructions somewhere I would very much appreciate that.

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1 –  Jim Feb 3 '13 at 20:07

3 Answers 3

up vote 2 down vote accepted

$H = \{f \in S(A): f(a_0) = a_0\}$ is the set of bijective functions in $S(A)$ such that $f$ maps $a_0$ to $a_0$.

Here, $a_0$ is a fixed element which every bijective function in $S(A)$ that maps $a_0$ to itself is a member. $H$ includes the identity map/function, but not all $f \in H$ are the identity function.

To prove $H$ is a subgroup of $S(A)$:

  • Clearly, $H$ includes as its identity $f_I$, the identity map of $S(A)$, since $f_I(a_0)$ is clearly $a_0$.

  • Now show that for any $f, g \in H$, $f\circ g \in H$: i.e., show that $H$ is closed under the function composition.

  • Show that for any $f \in H, f^{-1} \in H$. Hint: This must be the case, as every $f \in H$ is bijective, and if $f \in H$, then $f^{-1}$ exists, and since $f\in H$, $f(a_0) = a_0$ and $f^{-1}(f(a_0)) = f^{-1}(a_0) = f_I(a_0) = a_0$, since by definition, $f^{-1}\circ f = f_I$, so $f^{-1} \in H$

Then you will have shown $\,H\,$ is a subgroup of $\,S(A)$.

As for your second question (or aside), you might want to visit MathJax/LaTeX: a tutorial and "bookmark" it as a favorite!

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Hint: To check something is a subgroup, it suffices to show that $H$ is nonempty (this clear since the identity function is an element [why?]), the composition of two functions is an element of the group (also easy to argue since every function fixes $a_0$), and that the inverse of a function is an element of the group (which follows because each element is a bijection).

I leave the details to you, but hopefully with what is given above, you can work through this mostly on your own.

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The standard way to show that something is a subgroup is to show:

  1. It is nonempty
  2. If $f, g \in H$ then $fg^{-1} \in H$

I'll leave (1) to you and give you a hint for (2). If $f, g \in H$ then you know $f(a_0) = a_0$ and $g(a_0) = a_0$. Apply the function $fg^{-1}$ to $a_0$. The result should be $a_0$ and once you've shown this you can conclude that $fg^{-1} \in H$.

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