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Assume that $(G,*)$ and $(H,o)$ are groups and that $f:(G,*) \rightarrow (H,o)$ is a homomorphism.

Let $e_G$ and $e_H$ denote the identity elements of $G$ and $H$, respectively. Show that $f(e_G)=e_H$.

Approach: $f(e_G)=f(a*a^{-1})$ for $a,a^{-1} \in G$, so $f(a*a^{-1})=f(a)of(a^{-1})$.

If that’s true, then how do we know that $f(a^{-1})$ is the inverse of $f(a)$?

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    $\begingroup$ You don't, that would require $f(e)=e$. But try your same argument but with $a=e$ and it will work, we know $e^{-1}=e$ right. $\endgroup$ Commented Aug 1, 2016 at 4:13
  • $\begingroup$ I think you can carry on with $f(a)f(a^{-1}) = f(a)f(a)^{-1} = e_H$. If you are allowed to use $f(a^{-1}) = f(a)^{-1}$ $\endgroup$
    – IAmNoOne
    Commented Aug 1, 2016 at 4:26
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    $\begingroup$ @Nameless, that's what I have to prove next $\endgroup$ Commented Aug 1, 2016 at 4:26
  • $\begingroup$ @TheMathNoob, well I guess you can't use what I suggested. You need help with that one or...? $\endgroup$
    – IAmNoOne
    Commented Aug 1, 2016 at 4:27
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    $\begingroup$ @TheMathNoob, well actually we can more or less use what did here. Since you know $f(a)f(a^{-1}) = e_H$, what must $f(a^{-1})$ be? Actually I think my initial comment was backwards. $\endgroup$
    – IAmNoOne
    Commented Aug 1, 2016 at 4:30

3 Answers 3

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Using the fact that $e_G\ast e_G=e_G$ yields $$ f(e_G)=f(e_G\ast e_G)=f(e_G)\circ f(e_G)$$

Now multiply both sides by the inverse of $f(e_G)$ to obtain $f(e_G)=e_H$.

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  • $\begingroup$ ok so $f(e_G)=f(e_G) \circ f(e_G)$, $f(e_G) \circ f^{-1}(e_G)=(f(e_G)\circ f(e_G)) \circ f^{-1}(e_G)$ and hence $\circ $ is associative we can say $f(e_G)=e_H$ $\endgroup$ Commented Aug 1, 2016 at 4:22
  • $\begingroup$ That's correct. $\endgroup$ Commented Aug 1, 2016 at 4:23
  • $\begingroup$ That trick seemed to be very useless XD. $\endgroup$ Commented Aug 1, 2016 at 4:24
  • $\begingroup$ In other words: An element of a group is idempotent iff it is the neutral element. It is immediate that $f(e_G)$ is idempotent. $\endgroup$ Commented Jan 19, 2023 at 15:23
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The answers provided here are excellent, but here is something to just widen your mind for diversity.

So if $a\in G$ $$f(a) = f(a* e_G) = f(a)\circ f(e_G).$$

Left cancel by $f(a)^{-1}$ to get $e_H$

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  • $\begingroup$ Oh woahhh XDDDDD $\endgroup$ Commented Aug 1, 2016 at 4:37
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We have $e_G = e_G \ast e_G$, then

$$f(e_G) = f(e_g \ast e_G) = f(e_G)\circ f(e_G).$$

Now, $$e_H = [f(e_G)]^{-1}\circ f(e_G) = [f(e_G)]^{-1}\circ (f(e_G)f(e_G)) = f(e_G).$$

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