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The group $G$ is a finite group, a group with finite number of elements, and $H\triangleleft G $a normal subgroup. How can we prove that the index $|G/H|=|G|$ iff $H=\{e\}$, the identity element?

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Probably an overkill, but since the function $f: G \to G/H$ is an onto morphism, you can prove that it is injective if and only if $G$ and $G/H$ have the same number of elements... – N. S. Jan 31 '13 at 3:48
This result does not hold in general. See this question. A group $G$ where there exists some non-trivial $H$ such that $G/H\cong G$ is called non-Hopfian. – user1729 Jan 31 '13 at 11:12
thank you very much for your kind feedback – Faye Jan 31 '13 at 12:09
up vote 2 down vote accepted


You need to know your notation and definitions, that's about all there is to it.

Proofs are usually first and foremost about unpacking definitions, and knowing the notation.

What does it mean for $H \triangleleft G$ to be normal in $G$?

  • Answer: Yes, confirming your answer below,

    "A subgroup H of a group G is normal in G if and only if the left cosets equals the right cosets of $H$ in $G$ i.e. $gH=Hg$.

What does $[G:H]$, the index of $H \in G$, represent? And what does this mean with respect to $|G/H|$?

  • "... So the index represents the total number of cosets of $H$ in $G$."

    Yes, since $H$ is a normal subgroup of $G$, we can form the quotient group $G/H$ which contains all cosets of $H$ in the group $G$, so $[G: H] = |G/H|.$

And what must follow if $|G/H| = |G|$? I.e. What must the order of $H$ be for $|G/H| = |G|$ to be true?

  • "... the order of $H$ should be $1$ for the equation to hold.":
    Yes indeed.

And what is the only normal subgroup of $G$ of that order (it's the only subgroup, period, of that order)?

  • " that means the only normal subgroup in $H$ is $H$ itself?"

    Well, yes, $H$ is normal, as given: it means the only normal subgroups $H \triangleleft G$ such that $|G/H|= |G| \implies |H| = 1\;$ is $H = \{e\}$, the trivial (normal) subgroup.
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A subgroup H of a group G is normal in G if and only if the left coset equals the right coset. (gH=Hg) So the index represents the total number of cosets. And the order of H should be 1 for the equation to hold, so that means the only normal subgroup in H is H itself? – Faye Jan 31 '13 at 3:45
It means that the normal subgroup H of G must be of order 1, since the number of cosets of $H$ is equal to the order of $G$, that is $H$ must be the trivial (and normal) subgroup $H = \{e\}$. – amWhy Jan 31 '13 at 3:47
Ok yes, thank you for your help! – Faye Jan 31 '13 at 3:50

As @Jacob noted above, the proof is very short and it looks so clear. Since $H$ is a normal subgroup of $G$, so we can speak about a new group called Quotient Group $G/H$ which contains all right cosets of $H$ in the group. According to the definition, the index of $H$ in $G$ is denoted by $[G:H]$ and if the mother group is finite, you are allowed to write it as $$[G:H]=\frac{|G|}{|H|}$$ Here, $G$ is finite, so $|G|<\infty$ and then we have $$[G:H]=\frac{|G|}{|H|}=|G|\Longleftrightarrow H\cong\{e\}$$

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thank you also for your feedback... – Faye Jan 31 '13 at 12:57

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