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Let $\phi$ be a homomorphism of a group G into a group G'.
If $e =$ the identity element in G, then $\phi(e) =$ the identity element in G'.

Is this what Sharkos is trying to answer:
About homomorphisms, you only know $\phi(a) \phi(b) = \phi(ab)$.
ab is more complicated than anything we want to think about, hence just presuppose $b = e$.
Then $ \begin{align} \phi(a)\phi(e) & = \phi(a\color{magenta}{e}) \\ & = \phi(a) \end{align} $
Left multiply the last equation by $\color{green}{\phi(a)^{-1}}$: $\quad \phi(e) = id_{G'}.$

(1.) How do you predestine to rewrite $a$ as $ a = a\color{magenta}{e}$?
Or to 'presuppose $b = e$' ? I understand neither tricks.

(2.) What's the intuition?

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  • $\begingroup$ Hi Frank, do you have multiple accounts? Your habits strongly resemble those of another user. Just curious. Regards, $\endgroup$ – Potato Jan 26 '14 at 9:05
  • $\begingroup$ @Potato - Searching for "envisage and envision" seems like the easiest way to identity the OP :) $\endgroup$ – Sharkos Jan 26 '14 at 10:50
  • $\begingroup$ These are the certain kinds of tricks that you have to practice and get used to. Once you get used to it, it will become second nature and will help you in the long run. I recommend working on lots of problems and getting good at solving the problems/theorems that you already know. $\endgroup$ – user357980 Mar 3 '18 at 3:19
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Since $$\phi(e)=\phi(ee)=\phi(e)\phi(e),$$ you can cancel one of the $\phi(e)$ in $G'$, then leaving you with $e'=\phi(e)$.

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  • $\begingroup$ did you know that in a group $ab=ac$ implies $b=c$ ? $\endgroup$ – janmarqz Jan 22 '14 at 20:59
  • $\begingroup$ @Frank Muer: i would like to understand why such a brief argument is not preferred $\endgroup$ – janmarqz Jan 22 '14 at 21:47
  • $\begingroup$ Thanks a lot. Upvoted. I actually like easier proofs too, but the proof in the question is Fraleigh's. I don't know why Fraleigh didn't prefer your argument. Can you please get back to me on my update in your answer? Please don't write as a comment. $\endgroup$ – Group Theory Jan 26 '14 at 7:53
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    $\begingroup$ @FrankMuer, from the Fraleigh's ages to now there were some evolution :D $\endgroup$ – janmarqz Jan 26 '14 at 17:35
  • $\begingroup$ @janmarqz exactly what I needed. $\endgroup$ – Karl Mar 30 '15 at 22:19
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  1. The identity element $\iota$ in a group $H$ has the property (by definition) that $hh^{-1} = \iota$ for all $h\in H$. Therefore, in order to find out what the identity $e'$ of $G'$ is, one could write down $e' = \phi(a) ^{-1} \phi(a)$. This is essentially what they do. You could choose a specific $a$ if you wanted; for example, as janmarqz suggests you can use $a=e$.

  2. By the looks of it, the only rules you know about $\phi$ are that $\phi(a) \phi(b) = \phi(ab)$. The simplest way to use this is to observe that because $a = ae$, we have $\phi(a) = \phi(ae) = \phi(a)\phi(e)$ - in other words to choose $b=e$. (Again, we could have chosen $a=e$ to make things even simpler.)

  3. The intuition is "If $e$ doesn't do anything to an element of $G$, and $\phi(a)$ does the same sort of thing in $G'$ as $a$ does in $G$, then $\phi(e)$ must not do anything to an element of $G'$. Therefore it must be the identity."


Edit: Choosing a specific $a\in G$ is unnecessary so long as one is sure that it is possible to choose some $a\in G$. (To prove "There are two natural numbers whose difference is 0", one can just say "Take any number $n \in\mathbb N$; consider $n-n=0$. We are done", and not choosing some specific $n$ makes no difference.)

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  • $\begingroup$ By the way, re: the previous edit which was made - I used distinct generic letters for that part because one actually applies it to two cases, namely $a \in G$ and $\phi(a) \in G'$. I've emphasized this by making it more distinct. $\endgroup$ – Sharkos Jan 26 '14 at 10:56
  • $\begingroup$ As far as I can tell I answered the beginning of the thing labeled 1 under your update already at the start of my answer, and the second half in my update. I answered 2 in my original response. $\endgroup$ – Sharkos Feb 4 '14 at 8:31
  • $\begingroup$ Do you understand my example proof with natural number $n$ which can be anything? It's left arbitrary just because it makes no difference to the difficulty of the proof but points out marginally more general facts along the way. $\endgroup$ – Sharkos Feb 4 '14 at 8:36
  • $\begingroup$ We have to find some way of using the fact that $\phi(a)\phi(b) = \phi(ab)$ because this is all we have to go on. But $ab$ is more complicated than anything we want to think about, so it's much simpler to just suppose, say, $b=e$ so one of them goes away. $\endgroup$ – Sharkos Feb 4 '14 at 12:38
  • $\begingroup$ Those are exactly the same, just in a different order. $\endgroup$ – Sharkos Feb 4 '14 at 16:10
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Without digging into any thing deep or complicated, you only have three pieces of knowledge:

  • $e'$ is the only element that ever satisfies an equation in $G'$ of the form $e'x = x$ or $ye' = y$
  • $e$ satisfies equations in $G$ of the form $ex = x$ and $ye = y$
  • $\phi(xy) = \phi(x) \phi(y)$ (and similarly for products of more than two things)

By the first point, if you want to prove $\phi(e) = e'$, the only option we have (if we want to use these simple pieces of knowledge) is to find an equation $\phi(e)x = x$ or $y \phi(e) = y$. Trying to figure out how to use the other pieces of information to do this leads to a proof.

The proof could also have been derived from a simple approach of "I have facts, I'll do something with them and see if the result is useful". e.g. looking at the second and third points and seeing that you can combine them... so you do so, and see if the result is useful. (and it turns out to be exactly what you need to use the first point to complete the proof)

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  • $\begingroup$ Can you please notify me about my updated question in your answer, and not in comments? $\endgroup$ – Group Theory Apr 27 '14 at 16:50

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