I'm just learning about isomorphisms. Suppose $f:G\rightarrow G'$ is an isomorphism from the groups $G$ to $G'$. Why then is the kernal of $G$ equal to $\{e_G\}$? According to my source, we have $$\text{Ker}(f)=\{g\in G:f(g)=e_{G'}\},$$ and for an isomorphism we always have $$\text{Ker}(f)=\{e_G\}.$$ I know that an isomorphism is a bijection and get the idea of a bijection, but can't yet fathom how this relates to the associated kernel.
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$\begingroup$ The kernel is a measure of non-injectivity. An isomorphism is injective, so its kernel is as small as possible, that means it's $\{e_G\}$, since $e_G$ always is in the kernel. $\endgroup$– Daniel FischerNov 18, 2013 at 8:33
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7$\begingroup$ Note that the kernel is a property of $f$, not $G$. $\endgroup$– goblin GONENov 18, 2013 at 8:33
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$\begingroup$ Part of the definition of "bijection" is "one-one". Since the identity is always in the kernel, there's no room there for anything else in a one-one map. $\endgroup$– Gerry MyersonNov 18, 2013 at 8:33
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1$\begingroup$ @user18921 should I be speaking of $\text{Ker}(f)$ instead of $\text{Ker}(G)$ ? $\endgroup$– pshmath0Nov 18, 2013 at 8:42
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1$\begingroup$ @pbs, yes absolutely. $\endgroup$– goblin GONENov 18, 2013 at 8:48
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2 Answers
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Because an isomorphism is a bijection, every element of $G'$ has exactly one preimage; in particular there is exactly one preimage of $e_{G'}$. Since $e_G$ is already a preimage of $e_{G'}$, that is the one. The kernel consists of all preimages of $e_{G'}$, so is just $\{ e_G \}$.
Or, if you prefer the reasoning in symbols, $$g \in \text{Ker}(f) \iff f(g) = e_{G'} \iff f(g) = f(e_G) \iff g = e_G.$$ The last $\iff$ uses the assumption that $f$ is a bijection (actually, just that it is injective, as that is enough for the kernel to be trivial).
answered Nov 18, 2013 at 8:33
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$\begingroup$ Thanks, I think I understand, except for why is it we can assume that "...$e_G$ is already a preimage of $e_{G'}$" ? Is this to do with the structure preserving property $f(g\cdot h)=f(g)\cdot f(h)$, where the $\cdot$ in the left is multiplication in $G$ and the $\cdot$ in the right is multiplication in $G'$ ? $\endgroup$– pshmath0Nov 18, 2013 at 9:06
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1$\begingroup$ $f(e_G)=f(x\cdot x^{-1})=f(x)\ast f(x^{-1})=f(x)\ast {f(x)}^{-1}=e_{G'}$. You can prove by exercise that $f(x^{-1})={f(x)}^{-1}$ if you don't know it. $\endgroup$– DubiousNov 18, 2013 at 9:15
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$\begingroup$ @Galoisfan Thanks, that makes sense now :-) $\endgroup$– pshmath0Nov 18, 2013 at 9:20
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3$\begingroup$ Sorry, the statement in the last comment requires that $f(e_G)=e_{G'}$; in particular $f(e_G)=e_{G'}$ implies that $f(x^{-1})={f(x)}^{-1}$. A better proof should be the following: $f(e_G)=f(e_G\cdot e_G)=f(e_G)\ast f(e_G)$, so by the right cancellation you have that $f(e_G)=e_{G'}$. $\endgroup$– DubiousNov 18, 2013 at 9:25
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The "size" of the kernel measures the "injectivity" of a function in the following sense:
Proposition: $f$ is injective if and only if ker$(f)=\{e\}$.
If ker$(f)=\{e\}$, from the equality $f(x)=f(y)$, you have that $f(xy^{-1})=e'$ i.e. $xy^{-1}\in$ ker$(f)$. From the hypothesis, $xy^{-1}=e$, so you can conclude that $x=y$.
Viceversa suppose that $f$ is injective; if $x\in$ ker$(f)$, $f(x)=e'=f(e)$ and follows $x=e$ thanks to the injectivity .
