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If $H$ is a normal subgroup of $G$ and $|H| = 2$, show that $H \subseteq Z(G)$.

Proof:

Since $|H| = 2$, we have $H = \{e, a\}$, where $a \in G$ is such that $|a| = 2.$ Let $g \in G$ be arbitrary. Then $gHg^{−1} = \{geg^{−1}, gag^{−1}\} = \{e, gag^{−1}\}$. Since $H$ is a normal subgroup of $G$, we have $gHg^{−1} = H$, that is $\{e, a\} = \{e, gag^{−1}\}$. Hence, $a = gag^{−1}$, and so $ag = ga$. Since $g \in G$ was arbitrary, this means that $a \in Z(G)$. Also, obviously $e ∈ Z(G)$. Therefore $H = \{1, a\} \subseteq Z(G)$, as required.

My question:

Why is the line "Since $H$ is a normal subgroup of $G$, we have $gHg^{−1} = H$" true? Should it not read "Since $H$ is a normal subgroup of $G$, we have $gHg^{−1} \subseteq H$" ?

user26857
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IgnorantCuriosity
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2 Answers2

2

If your doubt is:

$\forall g \in G: gHg^{-1} = H \iff \forall g \in G: gHg^{-1} \subseteq H$

this is easily remedied.

One implication is clear, if $gHg^{-1} = H$ for any $g \in G$, then certainly $gHg^{-1} \subseteq H$, for clearly, $H \subseteq H$.

What usually isn't immediately apparent, is that the seemingly weaker condition:

$gHg^{-1} \subseteq H$ (for all $g \in G$) actually implies the seemingly stronger condition:

$gHg^{-1} = H$ (for all $g \in G$). The "for all $g \in G$" part is very important.

For suppose $h \in H$ is arbitrary. We may re-write this as:

$h = ehe = ehe^{-1} = (gg^{-1})h(gg^{-1})^{-1} = g(g^{-1}hg)g^{-1}$.

Since $g^{-1} \in G$ whenever $g$ is, and by supposition we have $g^{-1}Hg \subseteq H$, then $g^{-1}hg = h'$, for some $h' \in H$.

Thus $h = g(g^{-1}hg)g^{-1} = gh'g^{-1}$, which is an element of $gHg^{-1}$.

This shows that $H \subseteq gHg^{-1}$, so that (along with $gHg^{-1} \subseteq H$) we obtain $gHg^{-1} = H$,

But in the proof you are reading, we don't even need the stronger condition, merely assuming $gHg^{-1} \subseteq H$ will suffice for the proof.

For if $\{e,gag^{-1}\} \subseteq H = \{e,a\}$, we have only two possibilities for what $gag^{-1}$ might be: either $e$, or $a$.

If $gag^{-1} = e$ this leads to $a = g^{-1}eg = g^{-1}g = e$, a contradiction, since $a$ is assumed distinct from $e$ (or else $|H| = 1$, and not $2$). So, we are forced to concede that $gag^{-1} = a$, and the conclusion follows.

David Wheeler
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1

One can actually show that "$gHg^{-1} = H$ for any $g$" is an equivalent definition for $H$ being a normal subgroup. See here for example.

angryavian
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