# Complex character of a finite group $G$

We let $$G$$ be a finite group.

If $$\chi$$ is a complex character of $$G$$, we define $$\overline{\chi}:G \to \mathbb{C}$$ by $$\overline{\chi}(g)=\overline{\chi(g)}$$ for all $$g \in G$$, and define $$\chi^{(2)}:G \to \mathbb{C}$$ by $$\chi^{(2)}(g) = \chi(g^2)$$. We write $$\chi_{S}$$ and $$\chi_{A}$$ for the symmetric and alternating part of $$\chi$$. We note that $$\chi_{S}$$ and $$\chi_{A}$$ are characters of $$G$$ with $$\chi^{(2)}=\chi_{S} + \chi_{A}$$ and $$\chi^{(2)}=\chi_{S} + \chi_{A}$$.

First, I want to show that $$\overline{\chi}$$ is a character of $$G$$. Now, we can show that $$\overline{\chi}(g)=\overline{\chi(g)}=\chi(g^{-1})$$ for all $$g \in G$$ thus $$\overline{\chi}(g)$$ is a character. Is that OK?

Next, I want to show that $$\chi$$ is irreducible iff $$\overline{\chi}$$ is irreducible.

For $$(\implies)$$ we assume that $$\overline{\chi}$$ is not irreducible. Thus we must have a reducible representation $$\rho:G \to GL(V)$$. But then $$\chi$$ must also be reducible, w.r.t. to that reducible representation, which is a contradiction. $$(\impliedby)$$ we can show by the same argument. Tbh, it doesn't seem correct to me, I don't think that I really understand what could go wrong here.

Lastly, we let $$\chi_{1}$$ be the trivial character of $$G$$. If I understand it correctly, $$\chi_{1}(g)=1$$ for all $$g \in G$$. We want to show that $$\langle \chi , \overline{\chi} \rangle= \langle \chi_{S},\chi_{1}\rangle + \langle \chi_{A}, \chi_{1} \rangle$$.

We have:

$$\langle \chi , \overline{\chi} \rangle = \frac{1}{|G|} \displaystyle\sum_{g \in G} \chi(g)\overline{\chi(g)} = \frac{1}{|G|} \displaystyle\sum_{g \in G} \chi(g)\chi(g^{-1})$$

and now I am not sure where to go from here, I'd appreciate any hints.

Recall that a character $$\chi$$ is irreducible if and only if $$\langle \chi,\chi\rangle=1$$. Note then that
$$\langle \overline{\chi},\overline{\chi}\rangle =\frac{1}{|G|}\sum_{g\in G}\overline{\chi}(g)\overline{\overline{\chi}(g)}=\frac{1}{|G|}\sum_{g\in G}\overline{\chi(g)}\chi(g)$$
$$\langle \chi,\chi\rangle=\frac{1}{|G|}\sum_{g\in G}\chi(g)\overline{\chi(g)}$$
Thus, we see that $$\langle \chi,\chi\rangle=\langle\overline{\chi},\overline{\chi}\rangle$$.
• Thank you! Somehow, I wasn't aware of this. Is the way of showing that $\overline{\chi}$ is a character correct? Apr 9, 2019 at 13:10
{"ft_lang_label":"__label__en","ft_lang_prob":0.75721824,"math_prob":1.0000036,"size":1678,"snap":"2022-05-2022-21","text_gpt3_token_len":568,"char_repetition_ratio":0.17502987,"word_repetition_ratio":0.016597511,"special_character_ratio":0.36412397,"punctuation_ratio":0.11461318,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000085,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-25T06:34:00Z\",\"WARC-Record-ID\":\"<urn:uuid:eb2235a6-8e29-4c9f-ae6b-204dc23338ac>\",\"Content-Length\":\"226917\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2e6e2588-f948-4763-884c-f61c5e2b3434>\",\"WARC-Concurrent-To\":\"<urn:uuid:e1a907c7-fb49-4e5b-a042-ecd4127e858b>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3180890/complex-character-of-a-finite-group-g\",\"WARC-Payload-Digest\":\"sha1:MWU4HAQRJFKNSKBNHROHLQVD4CHC6SDZ\",\"WARC-Block-Digest\":\"sha1:XXBNJB5HKITE7XTSBAF727OPKAQSFPUJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662580803.75_warc_CC-MAIN-20220525054507-20220525084507-00605.warc.gz\"}"}