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Let $H=\{(1), (1,2)(3,4)\}$ in $A_4$. Show $H$ is not Normal in $A_4$ using the definition of normal.

So I know that $A_4$ is the alternating group on $4$ letters, but I don't understand what that means. Can someone please explain? I also know that for a subgroup $H$ of $G$ to be normal, its left and right cosets must coincide. ie: $aH=Ha$. From here, how do I use this definition to prove that $A_4$ is not normal?

My plan was this: the left cosets are:

$1+H= \{(2),(2,3),(4,5)\}$ $\leftarrow$ Can you have $5$ in $A_4$, or would it be $0$ or $1$???

$2+H= \{(3),(3,4)\}$

$3+H= \{(4)\}$,

I think my main issue is that I do not understand what it means for something to be an "alternating group on 4 letters". Also, am I on the right track for trying to see if the cosets coincide?

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    $\begingroup$ There are some misunderstandings here. First of all, $H$ (at least in the way you wrote it) is not a group. I suspect that it was meant to be $\{(1),(1,2)(3,4)\}$, but this is just a guess. Then there seems to be some confusion as to what the multiplication is in this situation (are you familiar with the symmetric group and permutations?). $\endgroup$ Dec 15, 2014 at 6:01
  • $\begingroup$ Yes! I took another look and you are correct! $\endgroup$
    – Kaitlyn
    Dec 15, 2014 at 6:09

2 Answers 2

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Hint:

$$(1\;2\;3)\color{red}{(1\;2)(3\;4)}(1\;2\;3)^{-1}=(123)\color{red}{(12)(34)}(132)=(14)(23)\notin H$$

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Hint: That's a blunder! The cosets are formed under group operations. In permutation group, composition (or loosely, multiplication) and not addition is the group operation.

Open the link to know how a composition is evaluated.

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  • $\begingroup$ So rather than adding 1+H, 2+H etc to find the cosets, it would be multiplying: 0H,1H,2H,3H. would you multiply 4H as well, again, I am confused by exactly what it means for something to be "an alternating group on 4 letters". also, now that the question is changed im even more confused. $\endgroup$
    – Kaitlyn
    Dec 15, 2014 at 6:30
  • $\begingroup$ $aH$ is a coset if $a\in A_4$. Note that $0,1,2,3,4$ are not in $A_4$ $\endgroup$ Dec 15, 2014 at 6:37
  • $\begingroup$ so what is in A4? $\endgroup$
    – Kaitlyn
    Dec 15, 2014 at 6:40
  • $\begingroup$ I do not understand what A4 is.. I think that that is the main problem. $\endgroup$
    – Kaitlyn
    Dec 15, 2014 at 6:40
  • $\begingroup$ You should read, Permutation groups, then Alternating group. I recommend Gallian. $\endgroup$ Dec 15, 2014 at 6:43

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