I notice that conjugation works, but:
$(12)(123) = (123)(12) \rightarrow (1)(23)=(13)(2)$
Is clearly wrong. Doesn't normality imply that $hg=gh$ for $H$ being normal in G? What am i doing wrong here?
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3$\begingroup$ Another way of looking at Normal subgroups is the following: if H is a normal subgroup, then Hx=xH, ie. the left cosets are equal to the right cosets. This however does not imply that H is a Centralizer. $\endgroup$– EverstudentSep 2, 2020 at 22:42
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2$\begingroup$ Normality is akin to commutativity but at the level of subgroups. It is not the same however. $\endgroup$– Cameron WilliamsSep 2, 2020 at 22:43
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2$\begingroup$ H being normal does not imply that hc=ch. $\endgroup$– EverstudentSep 2, 2020 at 22:43
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2$\begingroup$ You can also argue that because $[S_3:A_3]=2$, $A_3$ is a subgroup of index $2$, so it is normal in $S_3$. $\endgroup$– ilovebulbasaurSep 2, 2020 at 22:48
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$\begingroup$ @ilovebulbasaur Why is it that a subgroup of index 2 has to be normal? Update: Ah, never mind, it's because it has only two cosets H and Hx, so Hx=xH. Anyway, that's an interesting argument. $\endgroup$– EverstudentSep 2, 2020 at 23:11
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1 Answer
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A normal subgroup H of G, is a subgroup such that for any $h$ from $H$, $xhx^{-1} \in H$ i.e. all the conjugates ($xhx^{-1}$) of $H$ are in $H$. Another way to look at it is the following: the left and right cosets of $H$ are the same, meaning $Hx=xH$. Basically saying that multiplying $H$ by an element is commutative. (Remember $Hx$ is the set all of elements of $H$ multiplied by $x$).
However this does not imply that the elements of $H$ commute with themselves or with other elements of $G$. There is another subgroup which is called the centralizer, by definition it's elements are those elements of $G$ which commute with all of the elements of $G$.
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$\begingroup$ Why the downvote? I am trying to explain things in detail and give more scope. $\endgroup$ Sep 2, 2020 at 22:50
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